“A topologist is someone who can’t tell a coffee-cup from a donut." Topology is concerned with how or where things are or aren’t joined up; whether or where there are breaks or holes; how things are connected. But what do we mean by ‘joined’ or ‘connected’? Intuitively, it’s something to do with being ‘as close as possible’, or ‘as close as we like’. But to say things are ‘close’ we need an idea of distance.
In Analysis I we very often used the distance \(|x-y|\) between two real numbers. We used inequalities involving distances to define limits, convergence, and continuity. In Complex Analysis II we looked briefly at metrics: A metric on a set gives a real value for any two elements of the set. These values must behave in certain ways like distances, but there are many metrics which are not the same as our familiar Euclidean distance. We could use any metric to define an open ball round a point, and using these open balls we defined open sets in general.
A main objective of Topology III, for most of the first term, is to show how we can skip over the idea of a metric, and start from open sets. If a collection of subsets of a set \(X\) satisfies certain requirements, we can call these subsets ‘open’, and the collection of open subsets ‘a topology on \(X\)’. The set \(X\) with its topology is then a ‘topological space’.
We then find ways to do analysis using open sets rather than a distance or metric. We redefine limits, convergence, continuity, connectedness, and compactness using open sets. Some of these new definitions are quite surprising.
So our topologist doesn’t care about the lengths (and angles) which, in traditional Euclidean geometry, define shapes. This is why they don’t distinguish between a coffee-cup and a donut: both have just one hole.
The last part of the term is more algebraic. In Algebra II we made quotient groups by dividing a group out by a normal subgroup H, which meant declaring the normal subgroup to be the identity. Similarly, in Chapter 6, we can make quotient spaces by reducing a whole subspace to a point. In Chapter 7 we start from groups, and give them a topology (that is, define their open sets) to get a topological group. We also recall the idea of a group acting on a set: here the set will be a topological space, the group will sometimes also be topological.
A metric space is a pair \((X,d)\) where \(X\) is a set and \(d\colon X\times X\to [0,\infty)\) is a function satisfying the following:
\(d(x,y)=0\) if and only if \(x=y\).
\(d(x,y)=d(y,x)\) for all \(x,y\in X\). (symmetry)
\(d(x,y)\leq d(x,z)+d(z,y)\) for all \(x,y,z\in X\). (triangle inequality)
We call \(d\) a metric on \(X\).
Let \(p \in [1,\infty)\). \({\mathbb R}^n\) has a metric \[d_p(x,y)=({ \vert x_1-y_1 \vert^p+\cdots + \vert x_n-y_n \vert^p})^{1/p} = \left( \sum_{i=1}^n \vert x_i-y_i \vert^p\right)^{\frac{1}{p}}\]
\({\mathbb R}^n\) also has the metric \[d_\infty(x,y) = \max\{|x_i-y_i|\,|\,i=1,\ldots,n\}.\]
Let \(C([a,b])=\{f\colon [a,b]\to {\mathbb R}\,|\,f \mbox{ is continuous}\}\). Then \[d(f,g)=\sup \{|f(x)-g(x)|\,|\,x\in [a,b]\}\] is a metric on \(C([a,b])\).
Let \(X\) be any set. Then \[d(x,y)=\left\{ \begin{array}{cl} 0 & \mbox{if }x=y \\ 1 & \mbox{if }x\not=y \end{array} \right.\] is a metric, called the discrete metric on \(X\).
The first example alone includes an uncountable infinity of different metrics, depending on the value of \(p\) (although for \(n=1\) they all become simply \(|x-y|\)). The case \(p=2\) gives our familiar Euclidean distance in \({\mathbb R}^n\). The case \(p=1\) is the taxicab metric. As \(p\) increases, more and more weight is given to the larger \(|x_i - y_i|\), at the expense of the smaller. This is why metric (2), which considers only the largest \(|x_i-y_i|\), is written as \(d_\infty\).
The set in (3) has continuous functions as its elements; evaluating these at different choices of \(x \in [a,b]\) is like looking at different co-ordinates of a point in \({\mathbb R}^n\). So taking the sup over all \(x\) in (3) is like taking the max over all \(i\) in (2).
The last example is peculiar. There is no idea of two elements being ‘close’ or ‘far’; they are simply ‘the same’ or ‘different’.
Examples 1.2 are all metric spaces.
Proving in general that \(d_p\) satisfies (c), the triangle inequality, is surprisingly awkward.1 But for \(p_1\) and \(p_\infty\) it’s quite easy (see Q ??). We will check the other two examples here.
For Example (3), property (a) is easy to show: \[\begin{aligned} d(f,g) =0 & \Leftrightarrow & \sup \{|f(x)-g(x)|\,|\,x\in [a,b]\} =0 \\ & \Leftrightarrow & f(x) = g(x) \;\; \forall x \in [a,b] \\ & \Leftrightarrow & f=g \end{aligned}\]
Symmetry follows easily from the fact that \(|x-y|\), our usual metric on \({\mathbb R}\), is also symmetric, so \(|f(x)-g(x)| = |g(x) -f(x)|\). Similarly, (c) follows from the triangle inequality for \(|x-y|\), but given we functions \(f, g \in C[a,b]\) we must start by finding a \(c \in [a,b]\) such that \(|f(c)-g(c)| = \sup \{|f(x)-g(x)|\,|\,x\in [a,b]\}\). We know that there is such a \(c\), because \([a,b]\) is closed and bounded, and \(f,g\) and so also \(f-g\) are continuous, so the sup must be attained. Then, for any function \(h \in C[a,b]\) we have: \[\begin{aligned} d(f,g) = |f(c)-g(c)| & \leq & |f(c)-h(c)|+|h(c)-g(c)| \\ & \leq & \sup \{|f(x)-h(x)|\,|\,x\in [a,b]\} + \sup \{|h(x)-g(x)|\,|\,x\in [a,b]\} \\ & = & d(f,h) + d(h,g) \end{aligned}\] For example (4), properties (a) and (b) follow straight from the definition. For the triangle inequality, the LHS can only take the value 0 and 1; the RHS, 0, 1, or 2. If the RHS is 1 or 2, the inequality is certainly satisfied. But if the RHS is 0, then \(x=z\) and \(z=y\), so \(x=y\) and the LHS is 0 also.
In any metric space, we can define an open or closed ball of any radius around any point. This idea should be familiar, though you may have used slightly different notation.
Let \((X,d)\) be a metric space, \(r>0\) and \(x\in X\). Then \[B(x;r)=\{y\in X\,|\,d(x,y)<r\}\] is called the open ball of radius \(r\) around \(x\), and \[D(x;r)=\{y\in X\,|\,d(x,y)\leq r\}\] is called the closed ball of radius \(r\) around \(x\).
If we are dealing with more than one metric, say \(d_1\) and \(d_2\), we can specify \(B_1(x;r)\) or \(B_1(x;r)\). Note the semicolon ‘;’ which helps to distinguish this notation from open and closed intervals.
The open and closed balls \(B(0;1)\) and \(D(0;1)\) are defined for all the metrics of Examples 1.2.
Remember, here \(1\in {\mathbb R}\), but 0 is a point, which must be suitably defined for your space. For metrics \(d_1, d_2\), and \(d_\infty\), it is easy to draw or imagine \(B(0;1)\) and \(D(0;1)\) for \(n=2\), or imagine them for \(n=3\). (Q ??). For the discrete metric, with \(X = {\mathbb R}^2\), we have \[\begin{aligned} B(0;1) & = & \{y\in X\,|\,d(0,y)<1\} = {0} \\ D(0;1) & = & \{y\in X\,|\,d(0,y)\leq 1\} = {\mathbb R}^2 \end{aligned}\] For Example 1.2 (3) we can define our point 0 as the zero function, \(0(x) = 0 \;\; \forall x \in [a,b]\). Then \[\begin{aligned} B(0;1) & = & \{f \in C[a,b]\;|\;d(0,f)<1\}\\ & = & \{f \in C[a,b]\;|\;\sup \{|f(x)-0(x)|\,|\,x\in [a,b]\} < 1\} \\ & = & \{f \in C[a,b]\;|\;|f(x)|<1\; \forall x\in [a,b]\} \end{aligned}\] and similarly, \[D(0;1) = \{f \in C[a,b]\;|\;|f(x)|\leq 1\; \forall x\in [a,b]\}.\]
We can use the open ball to define a much wider class of sets.
Let \(X\) be a metric space. A subset \(U\subset X\) is called open, if for every \(x\in U\) there is \(\varepsilon>0\) such that \(B(x;\varepsilon)\subset U\). A subset \(A\subset X\) is called closed, if \(X \setminus A\) is open.
So intuitively, points in an open set always have ‘elbow-room’ in the set; they are never right on the edge. Still intuitively, we might say that an open set does not include any of its boundary. Similarly, we might say that a closed set includes all of its boundary. But then what is a set that includes only some of its boundary? Apparently it is neither open nor closed, and indeed such sets exist. More surprisingly, we shall see that some sets are both open and closed.
Let \({\mathbb R}\) have the usual \(|x-y|\) metric. Then, considered as sets,
any single point \(\{x\} \in {\mathbb R}\) is not open.
any single point \(\{x\} \in {\mathbb R}\) is closed.
(1) For any \(\varepsilon>0\), consider \(y=x+ \varepsilon/2\). Then \(y \in B(x, \varepsilon)\), but \(y\not\in\) the original set \(\{x\}\). So it is not open. (2) Now consider any \(y \in {\mathbb R}\setminus \{x\}\), that is, \(y \not = x\). Let \(\varepsilon =|x-y|>0\). Then \(x \not \in B(y; \varepsilon)\); that is, \(y \in B(y; \varepsilon) \subseteq {\mathbb R}\setminus {x}\). So \({\mathbb R}\setminus \{x\}\) is open, so \(\{x\}\) is closed.
More generally, the same is true for single points in \({\mathbb R}^n\), using the usual Euclidean metric \(d_2\): they are closed and not open.
Suppose \(X\) is a metric space and \(x \in N \subseteq X\). We call \(N\) a neighbourhood of \(x\) if there exists an open set \(V \subseteq X\) with \(x \in V \subseteq N\).
Putting Definitions 1.6 and 1.8 together gives the following:
Corollary 1.1. For \(x\) in a metric space \(X\), \(N \subseteq X\) is a neighbourhood of \(x\) if and only if there is some \(\varepsilon >0\) such that \(x \in B(x;\varepsilon) \subseteq N\).
So a neighbourhood \(N\) of a point \(x\) can be open or closed or neither or both: it only has to include ‘elbow-room’ for \(x\).
Notice that open and closed balls were defined directly from the metric, before we had Definition 1.6. Fortunately it turns out that they do satisfy these new definitions, so that our choice of words is not misleading:
Open balls are open, and closed balls are closed.
Let \(B(x;r)\) be an open ball, with \(x \in X\) and \(r>0\), and suppose \(y \in B(x;r)\). We need to find an open ball round \(y\) within \(B(x;r)\). So let \(s = d(x,y)\). Then we know \(s<r\), and we can set \(\varepsilon = r-s>0\). Now we claim that \(B(y;\varepsilon)\subset B(x;r)\). To show this, we consider any \(z \in B(y;\varepsilon)\). Then \[\begin{aligned} d(z,x) & \leq & d(z,y) + d( y,x) \\ & < & \varepsilon + s\\ & = & (r-s) +s = r.\end{aligned}\] So \(z \in B(x;r)\), and we have \(y \in B(y;\varepsilon)\subset B(x;r)\) as required.
Now let \(D(x;r)\) be a closed ball. To show that it is closed, we must show that \(X \setminus D(x;r)\) is open. We consider some \(y \in X \setminus D(x;r)\), so we know that \(s=d(x,y) >r\). Then setting \(\varepsilon = s-r>0\), we claim that \(B(y;\varepsilon)\subset X \setminus D(x;r)\). Let \(z \in B(y;\varepsilon)\). Then \[\begin{aligned} d(z,x) & \geq & d(x,y) - d( y,z) \\ & > & s -\varepsilon\\ & = & s - (s-r) = r. \end{aligned}\] So \(z \not \in D(x;r)\), and we have \(y \in B(y;\varepsilon)\subset X \setminus D(x;r)\) as required. This means \(X \setminus D(x;r)\) is open, so \(D(x;r)\) is closed.
You can see how, though there are many similarities between these proofs, the indirect definition of a closed set makes the second proof more complicated. It’s also worth noticing how we get the version of the triangle inequality that we want. This always involves three values, here \(d(x,y), d(y,z)\), and \(d(z,x)\). Since any one of these is \(\leq\) the sum of the other two, it’s also true that any one of them is \(\geq\) the difference of the other two - either way round! In the ‘open’ proof, we must show that \(d(z,x) < r\), so we start with it being \(\leq\) the sum of other two. In the ‘closed’ proof we must show that \(d(z,x) > r\), so we start with it being \(\geq\) the difference of the other two, and put these the right way round to end up with \(+r\).
We next look at how the set operations \(\cap\) and \(\cup\) work with open and closed sets.
Let \(X\) be a metric space.
\(X\) and \(\emptyset\) are both open and closed.
an arbitrary union of open sets is open.
a finite intersection of open sets is open.
a finite union of closed sets is closed.
an arbitrary intersection of closed sets is closed.
(a) For \(x \in X\), we have \(x \in B(x; 1) \subseteq X\). But as there is no point \(x\) in \(\emptyset\), Definition 1.6 is vacuously satisfied. So both are open. But \(X \setminus X = \emptyset\), and \(X \setminus \emptyset =X\), so both have open complements, and are therefore closed.
(b) Let \(U_i \subset X\) be open for all \(i \in I\). Then if \(x \in \bigcup_{i \in I} U_i\) we have \(x \in U_{i_0}\) for some \(i_0 \in I\). So there is some \(\varepsilon >0\) such that \(a \in B(x;\varepsilon) \subseteq U_{i_0} \subseteq \bigcup_{i \in I} U_i\). Hence \(\bigcup_{i \in I} U_i\) is open.
(c) Let \(U_i \subset X\) be open for \(i = 1, \ldots, n\). Then if \(x \in \bigcap_{i=1}^n U_i\) we have \(x \in U_i\) for \(i = 1, \ldots, n\). So for each \(i\) there is some \(\varepsilon_i >0\) such that \(x \in B(x;\varepsilon_i) \subseteq U_i\). Let \(\varepsilon = \mathop{\mathrm{min}}\{\varepsilon_1, \ldots \varepsilon_n\} >0\). Then for \(i = 1, \ldots, n\) we have \(x \in B(x;\varepsilon) \subseteq B(x;\varepsilon_i) \subseteq U_i\). Hence \(x \in B(x;\varepsilon) \subseteq \bigcap_{i=1}^n U_i\), as required.
(d) Let \(V_i \subset X\) be closed for \(i = 1, \ldots, n\). Then by definition \(X \setminus V_i \subseteq X\) is open for \(i = 1, \ldots, n\), so by (c) we know \(X \setminus (\bigcup_{i=1}^n V_i) = \bigcap_{i=1}^n (X \setminus V_i)\) is open, so \(\bigcup_{i=1}^n V_i\) is closed as required.
(e) Similar to (d), but use (b) rather than (c) - see Q??
The distinction between finite and arbitrary unions and intersections is important: notice that in \({\mathbb R}^2\) with the usual metric, \(\{0\}=\bigcap_{i=1}^\infty B(0;\frac{1}{i})\) so an arbitrary intersection of open sets need not be open.
As usual, the discrete metric is a bit surprising:
If \((X,d)\) is discrete, then any \(A\subset X\) is both open and closed.
Consider any \(a \in A \subseteq (X,d)\). We have \(B(a; 1/2) = \{a\} \subseteq A\). So every subset of \(X\) is open. But then also for any \(A \subseteq X\), \(X \setminus A\) is also open, so \(A\) is closed.
The familiar definition of convergence of a sequence can be used with any metric:
Let \((X,d)\) be a metric space.
A sequence in \(X\) is a function \(a\colon {\mathbb N}=\{0,1,2,3,\ldots\} \to X\). We write \((a_n)_{n\in {\mathbb N}}\) for the sequence, where \(a_n=a(n)\).
Let \(a\in X\). Then the sequence \((a_n)_{n\in{\mathbb N}}\) converges to \(a\), if for all \(\varepsilon>0\) there is \(n_0\in {\mathbb N}\) with \(d(a,a_n)<\varepsilon\) for all \(n\geq n_0\). We write \(\lim\limits_{n\to\infty}a_n=a\) or \(a_n\to a\).
In (b), the metric is implied. To be more explicit we could write \(a_n \stackrel{d}{\longrightarrow} a\). There are then (at least) three different ways to define continuity, which should also be mostly familiar:
Suppose that \((X, d_X)\) and \((Y, d_Y)\) are metric spaces, \(a \in X\), and \(f: X \rightarrow Y\) is a function. Then the following are equivalent.
For all \(\epsilon > 0\) there exists \(\delta > 0\) such that \[d_X(a,x) < \delta \,\, \implies \,\, d_Y(f(a), f(x)) < \epsilon {\rm ,}\]
For all \((a_n)\) with \(a_n \stackrel{d_X}{\longrightarrow} a\), we have \(f(a_n) \stackrel{d_Y}\longrightarrow f(a)\).
For all open \(U \subseteq Y\) with \(f(a) \in U\), we have that \(f^{-1}(U)\) is a neighbourhood of \(a\).
We shall prove the implications (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a).
(a) \(\Rightarrow\) (b): Assume (a) holds, and we have a sequence \((a_n)\) in \(X\), \(a_n \stackrel{d_X}{\longrightarrow} a\). We must show that \(f(a_n) \stackrel{d_Y}\longrightarrow f(a)\), that is, for all \(\varepsilon > 0\) there is some \(n_0 \in {\mathbb N}\) such that for \(n \geq n_0\) we have \(d_Y(f(a_n) ,f(a)) < \varepsilon\). So let \(\varepsilon >0\) be given. Then by (a) there is some \(\delta >0\) such that if \(d_X(a_n, a) < \delta\) then \(d_Y(f(a_n), f(a)) < \varepsilon\). But also, since \(a_n \stackrel{d_X}{\longrightarrow} a\), there is some \(n_0\in {\mathbb N}\) such that for all \(n\geq n_0\) we have \(d_X(a_n,a)<\delta\). Hence, for \(n\geq n_0\) we have \(d_Y(f(a_n) ,f(a)) < \varepsilon\) as required.
(b) \(\Rightarrow\) (c), by contrapositive: Assume that (c) does not hold. So there is some open set \(U\) with \(f(a) \in U \subseteq X\) such that \(f^{-1}(U)\) is not a neighbourhood of \(a\). This means that, for this \(U\), there is no open set \(V\) with \(a \in V \subseteq f^{-1}(U)\). Some obvious candidates for such \(V\) are the open balls \(B_X(a; 1/n)\), but by our assumption these do not satisfy: for all \(n\geq 1\), \(B_X(a; 1/n) \not\subseteq f^{-1}(U)\). Each \(B_X(a; 1/n)\) ‘sticks out’ somewhere from \(f^{-1}(U)\).
To show (b) is false, we need a sequence, and we make it by picking a point from each sticking-out region: for all \(n\), let \(a_n \in B_X(a; 1/n) \setminus f^{-1}(U)\). Since \(d_X(a_n,a) <1/n\), this sequence converges to \(a\). But also, since \(U\) is open there is some \(\varepsilon > 0\) such that \(B_Y(f(a), \varepsilon) \subseteq U\). Hence, since \(f(a_n) \not\in U\), we must have \(d_Y(f(a_n), f(a)) \geq \varepsilon\) for all \(n\). Thus \(f(a_n)\) does not converge to \(f(a)\), so (b) is false.
(c) \(\Rightarrow\) (a): Assume (c) holds, and let \(\varepsilon > 0\). Then \(B_Y(f(a); \varepsilon)\) is open, so by (c) \(f^{-1} (B_Y(f(a); \varepsilon))\) is a neighbourhood of \(a\). Then by Cor.1.9, for some \(\delta >0\) we have \(a \in B_X(a;\delta) \subseteq f^{-1} (B_Y(f(a); \varepsilon))\). This says that if \(d_X(a,x) < \delta\) then \(d_Y(f(a),f(x)) < \varepsilon\) as required.
If a function \(f\) as above satisfies property (a), we say it is continuous at point \(a \in X\).
If \(f: X \rightarrow Y\) is continuous at all points \(a \in X\) we say that \(f\) is continuous.
Of course, since we know that properties (a), (b), and (c) of Prop 1.14 are equivalent, we could have made Definition 1.15 use (b) or (c) instead. More interestingly, continuity (over the whole set) is equivalent to a property which is similar to (c), but not concerned with individual points, only with open sets.
A function \(f : X \rightarrow Y\) is continuous if and only if \(f^{-1}(U)\) is open for all open \(U \subseteq Y\).
To prove this we use the (c) version of Defn. 1.15: \(f\) is continuous if and only if we have property (c) for any point \(a \in X\).
\(\Rightarrow:\) Consider any open set \(U \subseteq Y\). Since \(f\) is continuous, for any \(a \in f^{-1}(U) \subseteq X\) (so \(f(a) \in U \subseteq Y\)) we have that \(f^{-1}(U)\) is a neighbourhood of of \(a\). It follows (by Cor. 1.9) that there is some \(\varepsilon >0\) such that \(a \in B_X(a;\varepsilon) \subseteq f^{-1}(U)\). But this means that \(f^{-1}(U)\) is open as required.
\(\Leftarrow:\) Consider any point \(a \in X\), and any open set \(U\) with \(f(a) \in U \subseteq Y\), so \(a \in f^{-1}(U)\). Our assumption is that \(f^{-1}(U)\) is open, so there is some \(\varepsilon >0\) such that \(a \in B_X(a;\varepsilon) \subseteq f^{-1}(U)\), as required for property (c), so \(f\) is continuous.
We are familiar with many functions \(f: {\mathbb R}\to {\mathbb R}\) that are continuous using the usual distance metric on \({\mathbb R}\).
Define \(I\colon C[0,1] \to C[0,1]\) by \[I(f) (x) = \int\limits_0^x f(t) dt.\] This is continuous with the metric from Example 1.2 (3).
Proving this is not really difficult, but it is confusing. See Q?? (with hints).
Again, we can end with the odd behaviour of the discrete metric:
Let us write \(d\) for the discrete metric, and \(d_2\) for the usual distance metric. Then
For any set \(X\), any function \(f\colon (X, d) \to ({\mathbb R}, d_2)\) is continuous.
But the identity function \(\mathop{\mathrm{id}}\colon ({\mathbb R},d_2)\to ({\mathbb R},d)\) is not continuous.
For both, we use Prop. 1.17. to test for continuity. (1): By Example 1.12 every subset of \(X\) is open. But then we are done, as \(f^{-1}(U)\) is always open. For (2), consider the set \(\{0\}\) which is open in \((X,d)\). Then \(f^{-1}(\{0\}) = \{0\}\), but a single point is not open in \((X, d_2)\) (see Example 1.7).
Topological spaces are a generalisation of metric spaces. We write down the properties satisfied by the collection of open sets in a metric space. Then, for any collection of subsets in a set, we can decide whether it has these properties, or not. If it does, we call these subsets ‘open’, the collection ‘a topology’, and the original set becomes a ‘topological space’.
Notation: For any set \(X\) we write \(\wp(X) = \{ A : A \subseteq X \}\) for the power set of \(X\), that is, for the set whose elements are all the subsets of \(X\).
Let \(X\) be a set. A topology on \(X\) is a subset \(\tau \subseteq \wp(X)\) which satisfies the following:
\(\emptyset\) and \(X\) are in \(\tau\).
If \(U_i\in \tau\) for all \(i\in I\), then \(\bigcup_{i\in I}U_i\in \tau\).
If \(U_1, U_2\in \tau\), then \(U_1\cap U_2\in \tau\).
The pair \((X,\tau)\) is called a topological space. The elements of \(\tau\) are called open subsets of \(X\). A subset \(A\subset X\) is called closed, if its complement \(X \setminus A\) is open. It is easy to show (by induction, Q??) that (c) is equivalent to:
If \(U_1, \ldots U_n \in \tau\), then \(\bigcap_{i=1}^n U_i\in \tau\).
Let \((M,d)\) be a metric space, and \(\tau_d\) the collection of open sets in the sense of Definition [def:open]. Then \((M,\tau_d)\) is a topological space. We say the metric \(d\) induces the topology \(\tau_d\).
Let \(X\) be any set, and \(\tau=\wp(X)\). This is called the discrete topology on \(X\).
Let \(X\) be any set, and \(\tau=\{\emptyset, X\}\). This is called the indiscrete topology on \(X\).
Let \(X=\{0,1,2,3,\ldots\}\) and \(\tau=\{\emptyset\}\cup \{U\subseteq X\,|\, (X \setminus U) \mbox{ is finite}\}\).
For example (1), Lemma 1:11 (a), (b), and (d) provide the properties of Definition 2.1 (a), (b), and (c’) respectively. For (2), since every set is in \(\tau\), all properties obviously hold; for (3), any unions or intersections of \(\emptyset\) and \(X\) are \(\emptyset\) or \(X\).
(4) takes a little more thought. Since \(X \setminus X = \emptyset\) which is finite, we have \(X \in \tau\), and \(\emptyset \in \tau\) has been specially arranged. For (b), either all \(U_i\) are \(\emptyset\), in which case so is \(\bigcup_{i\in I}U_i\), or some \(U_{i_0} \not = \emptyset\), in which case \[X \setminus \bigcup_{i\in I}U_i = \bigcap_{i\in I}(X \setminus U_i) \subseteq X \setminus U_{i_0},\] which is finite. Finally, for (c), if either \(U_1\) or \(U_2\) is \(\emptyset\) then so is their intersection; otherwise, \(X \setminus (U_1 \cap U_2) = (X \setminus U_1) \cup (X\setminus U_2)\), and the union of two finite sets is finite, so \((U_1 \cap U_2)\) is open as required.
Other properties from Lemma 1.11 also carry over from metric spaces to topological spaces:
Let \(X\) be a topological space. Then \(X\) and \(\emptyset\) are closed. Furthermore, arbitrary intersections of closed sets are closed, and finite unions of closed sets are closed.
The proofs are a straightforward exercise (Qs ??).
Let \((X,\tau)\) be a topological space, \(A\subset X\). Then \[\tau_A=\{A\cap U\,|\, U\in \tau\}\] is a topology, called the induced2 or subspace topology.
For (a), since \(\emptyset\) and \(X \in \tau\), \(\emptyset = A \cap \emptyset \in \tau_A\) and \(A = A \cap X \in \tau_A\). For (b), since \(\bigcup_{i\in I} U_i \in \tau\), we know \(\bigcup_{i\in I}(A \cap U_i) = A \cap \bigcup_{i\in I} U_i \in \tau_A\). And for (c) \((A\cap U_1) \cap (A \cap U_2) = A\cap (U_1 \cap U_2) \in \tau_A\).
Let \(\tau\) be the usual topology on \({\mathbb R}\) (that is, the one induced by the metric \(|x-y|\)). Then \(\tau_{{\mathbb Z}}\) is the discrete topology on \({\mathbb Z}\), but \(\tau_{{\mathbb Q}}\) is not the discrete topology on \({\mathbb Q}\). (Equivalently, we can say that \({\mathbb Z}\subset {\mathbb R}\) is discrete in the subspace topology, but \({\mathbb Q}\subset {\mathbb R}\) is not discrete in the subspace topology.)
The proofs are an exercise (Q??).
The metrics \(d_p\) for \(p \in [1, \infty)\) and also \(d_\infty\) all induce the same topology on \({\mathbb R}^n\).
This is called the standard topology on \({\mathbb R}^n\). It is perhaps surprising that, although different values of \(p\), or \(\infty\), give differently-shaped open balls (see Q ??), the topology (that is, the collection of all open sets) that they give is the same.
Rather than comparing for different values of \(p\), we shall show that for any \(p\), the metric \(d_p\) induces the same topology as \(d_\infty\). Recall that \[d_p(x,y)= \left( \sum_{i=1}^n \vert x_i-y_i \vert^p\right)^{\frac{1}{p}} \mathrm{ and }\;\; d_\infty(x,y) = \max\{|x_i-y_i|\,|\,i=1,\ldots,n\}.\]
First, we find two inequalities relating these metrics. Let \(0\leq m \leq n\) be the co-ordinate of maximum distance; that is, \(|x_m - y_m|\geq|x_i - y_i|\) for all \(0\leq i \leq n\). Then, since \(|x_i-y_i|\) is always \(\geq 0\), we have \[d_p(x,y)= \left( \sum_{i=1}^n \vert x_i-y_i \vert^p\right)^{\frac{1}{p}} \geq \left( |x_m-y_m|^p \right )^{\frac{1}{p}} = d_{\infty}(x,y).\] But also, \[d_p(x,y)= \left( \sum_{i=1}^n \vert x_i-y_i \vert^p\right)^{\frac{1}{p}} \leq \left(n |x_m-y_m|^p \right )^{\frac{1}{p}} = \sqrt[p]{n}\; d_{\infty}(x,y).\] Thus for any \(\varepsilon > 0\), \[d_p(x,y)< \varepsilon \Rightarrow d_{\infty}(x,y) < \varepsilon\;\;\;\; \mathrm{ and }\;\;\;\; d_{\infty}(x,y)< \frac{ \varepsilon}{\sqrt[p]{n}} \Rightarrow d_p(x,y) < \varepsilon.\] Using an obvious notation for open balls in each metric, it follows that \[B_p(x;\varepsilon) \subseteq B_{\infty} (x;\varepsilon) \;\;\;\; \mathrm{ and }\;\;\;\; B_{\infty}(x; \frac{\varepsilon} {\sqrt[p]{n}})\subseteq B_p (x;\varepsilon).\]
Now we are ready to consider general open sets in each topology. Suppose \(U\) is open in the topology induced by \(d_{\infty}\); that is, \(U \in \tau_{d_{\infty}}\). Then for any \(x \in U\) there is some \(\varepsilon\) such that \(x \in B_{\infty} (x;\varepsilon) \subseteq U\). But by the inclusions above it follows that \(x \in B_p (x;\varepsilon) \subseteq U\), so \(U \in \tau_{d_p}\) also. Conversely, if \(U \in \tau_{d_p}\), for any \(x \in U\) there is some \(\varepsilon\) such that \(x \in B_p (x;\varepsilon) \subseteq U\). But then we know that \(x \in B_{\infty}(x; \frac{\varepsilon} {\sqrt[p]{n}}) \subseteq U\), so \(U \in \tau_{d_{\infty}}\).
So far, our only non-trivial example of a topology not induced by a metric was Ex. 2.2 (4). We will find more of these, and we are also interested in properties which they may or may not have.
A topological space \(X\) is called Hausdorff, if whenever \(x,y\in X\) with \(x\not=y\), there exist open sets \(U,V\subseteq X\) with \(x\in U\), \(y\in V\) and \(U\cap V=\emptyset\).
You can think of this as saying the two points are ‘housed off’ from each other by the two disjoint open sets. This property seem intuitively obvious, and indeed our most familiar topological spaces do have it.
Let \(M\) be a metric space. Then \(M\) is a Hausdorff space.
(This is just a description of the obvious diagram.) Let \(x, y, \in M, x \neq y\), so \(d(x,y) = R > 0\). Set \(U = B(x, R/3)\) and \(V = B(y, R/3)\). (\(R/2\) would also work...) We must show that \(U\) and \(V\) are disjoint. So suppose not: there is some \(z \in U \cap V\). Then \[\begin{aligned} R =d(x,y) & \leq & d(x,z) + d(z,y) \\ & < & R/3 +R/3 = 2R/3\;\;\;\; \text{Contradiction.}\end{aligned}\] So \(U\cap V=\emptyset\) as required.
But there are topological spaces which are not Hausdorff.
Let \(X\) be a set with more than one element, and give it the indiscrete topology. Then \(X\) is not a Hausdorff space.
So here \(\tau = \{\emptyset, X\}\), and we have \(x, y, \in X, x \neq y\). The only open set containing \(x\) is \(X\), and the same is true for \(y\). But \(X\cap X = X \neq \emptyset\).
There are also topologies which are not induced by a metric, but are Hausdorff.
(Furstenberg’s topology on \({\mathbb Z}\)). A subset \(U \subseteq {\mathbb Z}\) is defined to be open if for every \(a \in U\) there exists a non-zero \(d \in {\mathbb Z}\) with \(a + d{\mathbb Z}\subseteq U\).
So in this topology, the role of the open ball in a metric space is played by sets \[a+ d{\mathbb Z}= \{ \ldots,\; a-2d,\; a-d,\; a,\; a+d,\; a+2d,\; \ldots \}\] - a shift of the multiples of \(d\). Any open set has this kind of ‘elbow-room’ round each of its points. We must show first that this is a topology, and then that it is Hausdorff.
We check each property of Definition 2.1: (a) \(\emptyset\) is vacuously open, and \({\mathbb Z}\) is open because for any \(a \in Z\) we can pick \(d=10\) (for example) and have \(a \in a+ 10{\mathbb Z}\subseteq {\mathbb Z}\). (b) Consider open sets \(U_i, i \in I\). If \(a \in \bigcup_{i\in I}U_i\), then \(a \in U_j\) for some \(j\in I\), so there is some \(d\) such that \(a+d{\mathbb Z}\subseteq U_j \subseteq \bigcup_{i\in I}U_i\). Hence \(\bigcup_{i\in I}U_i\) is open. (c) If \(a \in U_1 \cap U_2\) then there are \(d_1, d_2\) such that \(a+d_1{\mathbb Z}\subseteq U_1\) and \(a+d_2{\mathbb Z}\subseteq U_2\). But then \[a+d_1d_2{\mathbb Z}\subseteq a+d_1{\mathbb Z}\subseteq U_1,\] and similarly \[a+d_1d_2{\mathbb Z}\subseteq a+d_2{\mathbb Z}\subseteq U_2.\] So \(a+d_1d_2{\mathbb Z}\subseteq U_1 \cap U_2\), so this is open.
To show the topology is Hausdorff we take any \(a,b \in {\mathbb Z}, a \neq b\), so \(b-a \neq 0\). Then let \(U = a+ 2(b-a){\mathbb Z}\), and \(V= b+2(b-a){\mathbb Z}\). We need only to show these sets are disjoint. Suppose not: if \(z \in U \cap V\) then \(z = a+ 2(b-a)n = b+ 2(b-a) m\), for some \(m \neq n\) in \({\mathbb Z}\). This implies \(b-a = 2(b-a)(n-m)\), so \(n-m = 1/2\), which is a contradiction.
The idea of continuity transfers easily to topological spaces. Of course there are no \(\varepsilon\)s or \(\delta\)s; instead we use Prop1.14 (c) and Prop.1.17.
Let \(X,Y\) be topological spaces.
A function \(f\colon X\to Y\) is called continuous, if for every open subset \(V\subset Y\) the inverse image \(f^{-1}(V)\) is open in \(X\).
Let \(a\in X\). The function \(f\) is called continuous at \(a\), if for every open set \(V\subseteq Y\) with \(f(a)\in V\) there is an open set \(U\subseteq X\) with \(a\in U\) such that \(f(U)\subseteq V\).
We first need to check that the two parts of this definition are consistent:
A function \(f : X \rightarrow Y\) is continuous if and only if it is continuous at all points of \(X\).
As often, one direction is easy, one a bit more clever. \(\Longrightarrow\): Let \(a \in X\), and \(V\) be open, \(f(a) \in V \subseteq Y\). We are assuming (a), so \(f^{-1}(V)\) is open, and \(a \in f^{-1}(V) \subseteq X\). But as \(f(f^{-1}(V)) =V\), then \(f^{-1}(V)\) is the \(U\) required for (b). \(\Longleftarrow\): Let \(V \subseteq Y\) be open, and consider any \(a \in f^{-1}(V)\), so \(f(a) \in V\). We are assuming (b), so there is an open set \(U_a\) with \(a \in U_a \subseteq f^{-1}(V) \subseteq X\). Then notice that \[\bigcup_{a \in f^{-1}(V)}U_a \;\;\subseteq \;\;f^{-1}(V) \;\;= \bigcup_{a \in f^{-1}(V)}\{a\} \;\;\subseteq \bigcup_{a \in f^{-1}(V)}U_a.\] So \(f^{-1}(V) = \bigcup_{a \in f^{-1}(V)}U_a\) is a union of open sets, and so open as required for (a).
This definition of continuity generalizes all previously met ideas of continuity. These were (really) about continuity of maps between metric spaces, which is equivalent to continuity of maps with respect to the induced topologies. In practice this means that any functions that we would expect to be continuous are indeed so, by this new definition. Let us check a couple of examples of this:
Let \(X\) be a topological space and \(A\subseteq X\). The inclusion \(i\colon A \hookrightarrow X\) is continuous, if \(A\) is given the subspace topology.
Let \(U \subseteq X\) be open. Then \(i^{-1}(U) = U \cap A\), which is open in the subspace topology on \(A\), as required.
A composition of continuous functions is a continuous function.
We must show that, if \(f:X \rightarrow Y\) and \(g:Y \rightarrow Z\) are continuous, then \(g \circ f:X \rightarrow Z\) is continuous. Take \(U\) open in Z. Then since \(g\) is continuous \(g^{-1}(U)\) is open in Y, and so, since \(f\) is continuous, \((g \circ f)^{-1}(U) = f^{-1}(g^{-1}(U))\) is open in X as required.
Perhaps unexpectedly, closed sets are just as good for defining continuity.
Let \(f\colon X \to Y\) be a function between topological spaces \(X\) and \(Y\). Then \(f\) is continuous if and only if \(f^{-1}(A)\) is closed for every \(A\subseteq Y\) closed.
The proof is straightforward - see Q ??. Both directions use the fact that, for any set \(A \subseteq Y\), we have \(f^{-1}(A) = X \setminus f^{-1}(Y \setminus A)\).
These results can help us to determine whether sets are open or closed. We only have to define the sets using pre-images of continuous functions.
The \(n\)-sphere is the set \[S^n = \{ (x_1, \ldots,x_{n+1} \in {\mathbb R}^{n+1}\;|\;x_1^2 + \cdots x_{n+1}^2 = 1\}.\]
In the standard topology, the \(n\)-sphere is a closed subset of \({\mathbb R}^{n+1}\).
The function \(f:{\mathbb R}^{n+1} \rightarrow {\mathbb R}\) defined by \(f (x_1, \ldots,x_{n+1}) = x_1^2 + \cdots x_{n+1}^2\) is continuous, and the set \(\{1\}\) is closed (since \({\mathbb R}\setminus \{1\} = (-\infty, 1) \cup (1,\infty)\) is open). Since \(S^n = f^{-1}(\{1\})\), the pre-image of a closed set, it must be closed.
We can also consider sets of matrices. For \(n \in {\mathbb Z}_+\), the square \(n \times n\) matrices can be considered as a vector space: \(M_{n,n} \cong {\mathbb R}^{n^2}\). Using the standard topology on this space, we have:
\(GL_n({\mathbb R}) = \{ A \in M_{n,n} \;|\; \det(A) \neq 0\}\) is open.
\(SL_n({\mathbb R}) = \{ A \in M_{n,n} \;|\; \det(A) = 1\}\) is closed.
These sets are defined using the determinant function \(\det:{\mathbb R}^{n^2} \rightarrow {\mathbb R}\). For matrix \(A = (a_{i,j})\) this is defined by: \[\det (A) = \sum _{\sigma \in \rm{Sym}(n)} \left( \mathop{\mathrm{sgn}}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)} \right).\] This is a polynomial in the matrix entries \(a_{i,j}\), and so continuous. (We use the standard topology on \({\mathbb R}\), also.) Now since \(GL_n({\mathbb R}) = \det^{-1}({\mathbb R}\setminus \{1\}) = \det^{-1}( (-\infty, 0) \cup (0, \infty)))\) is the pre-image of an open set it is open, and since \(SL_n({\mathbb R}) = \det^{-1} (\{1\})\) is the pre-image of a closed set it is closed.