Using the related property for open sets, or otherwise, show that a map \(f:X \to Y\) of two metric spaces is continuous if and only if \(f^{-1}(F)\) is closed for all closed sets \(F\) in \(Y\).
Show that a map \(f:X \to Y\) between two metric spaces is continuous at \(x_0 \in X\) if and only if \[\lim_{n \to \infty} f(x_n) = f(x_0) \quad \text{ for every convergent sequence $\{x_n \}_{n \in \mathbb N}$ in $X$ with } x_n \to x_0.\] [Hint for ‘\(\Leftarrow\)’ direction: try a contrapositive argument. In particular, show that \(\exists \, \epsilon>0\) s.t. \(\forall \delta>0\) \(\exists \, x \in B_{\delta}(x_0)\) with \(f(x) \notin B_{\epsilon}(f(x_0))\), you can construct a sequence \(x_n \to x_0\) with \(f(x_n) \not\to f(x_0)\).]
Show that \(U:=\{ (x,y) \in \mathbb R^2: \: y-2x^2>0, \:\: y/x>2\}\) is open in \(\mathbb R^2\). [Hint: Use the properties of continuous functions. But be careful: e.g., \(y/x\) is not continuous on \(\mathbb R^2\) since it is not everywhere defined!]
Show that \(U:=\{(x,y) \in \mathbb R^2: \: (xy^2\sin(xy)>3) \text{ or } \; (e^{xy-2}+\log(x^2+1) < 3y)\}\) is open in \(\mathbb R^2\).
Show that \(U:=\{ (x,y) \in \mathbb R^2: \: xy^3/(xy-1) >2\}\) is open in \(\mathbb R^2\).
Suppose that \(X\) and \(Y\) are metric spaces with metrics \(d_X\) and \(d_Y\) and define a function \[d : (X \times Y) \times (X \times Y) \rightarrow [0,\infty)\] by \(d((x,y),(x',y')) = d_X(x,x') + d_Y(y,y')\). Check that \(d\) defines a metric on \(X \times Y\). Show that the map \(\pi : X \times Y \rightarrow X\) given by \(\pi(x,y) = x\) is continuous (with respect to the metrics \(d\) and \(d_X\)).
We call a map \(f:(X,d_1) \to (Y,d_2)\) between two metric spaces Lipschitz if there exists a positive constant \(C\) such that \[d_2(f(x_1),f(x_2)) \leq C d_1(x_1,x_2)\] for all \(x_1,x_2 \in X\). Show that if \(f\) is Lipschitz then \(f\) is continuous.
Let \(A \subseteq X\) be a non-empty subset of a metric space \(X\). We define the distance of a point \(\boldsymbol{x \in X}\) to \(\boldsymbol A\) by \[d(x,A): = \inf_{z \in A} d(x,z).\] Show that the function \(f: X \to \mathbb R: x \mapsto d(x,A)\) is Lipschitz and hence continuous. [Hint: Consider, for arbitrary \(z \in A\) and \(x,y \in X\), the inequality \(d(x,z) \leq d(x,y) + d(y,z)\) and take the infimum over \(z \in A\).]
Consider the map \(f:(-\pi,\pi] \to \{z \in \mathbb C: |z|=1\}\) given by \(f(t)=e^{it}\). Note that \(f\) is a bijection.
Show that \(f\) is continuous by using the \(\epsilon\)-\(\delta\) definition. (You may use the known facts that \(\sin\) and \(\cos\) are continuous on \(\mathbb R\).)
Prove separately that \(f\) is
continuous at \(t=\pi\) by showing that
the preimage \(f^{-1}(U_{\epsilon})\)
is open, where \(U_{\epsilon} =\{e^{it}: t \in
(\pi -\epsilon,\pi+\epsilon)\}\) for some \(\epsilon>0\)
[Note that due to the fact that \(f\) is \(2\pi\) periodic \(U_{\epsilon} =\{e^{it}: t \in (\pi
-\epsilon,\pi]\cup (-\pi,-\pi+\epsilon)\}\)].
However, show that the inverse map \(g=f^{-1}\) (the logarithm!) is not continuous on \(\{z \in \mathbb C: |z|=1\}\) by finding an open set \(V\) in \((-\pi,\pi]\) such that \(g^{-1}(V)\) is not open. [Compare this with Q9 below.]
Let \(f:X\to Y\) be a bijective continuous map of metric spaces and let \(X\) be compact. Show that the inverse map \(f^{-1}:Y \to X\) is continuous. [Hint: Use Q1 and the results from lectures concerning compactness.]
Notice that any complex number can be represented as \(z=re^{i\theta}\) for \(r \in \mathbb R\) and \(\theta \in (-\pi, \pi]\). Define a map \(f: \mathbb C\to \mathbb R\) by \(f(re^{i\theta}) = r / (\pi + \theta)\), for \(r \in \mathbb R\) and \(\theta \in (-\pi, \pi]\). Show \(f\) is not continuous at \(\pi\), but \(f|_L : L \rightarrow \mathbb R\) continuous for all straight lines \(L\) through the origin.
Discrete Sets
Let \(X\) be any metric space. We call
a subset \(A \subseteq X\)
discrete if for every point \(x \in A\) there is an open set \(U\) containing \(x\) that does not intersect any other point
of \(A\) (in other words, \(U \cap A = \{x \}\)).
Show that \(\mathbb Z\) is discrete inside \(\mathbb R\).
Show that \(\{ \tfrac{1}{n} : n \in \mathbb Z, n \not= 0\}\) is discrete in \(\mathbb R\), but \(\{ \tfrac{1}{n} : n \in \mathbb Z, n \not= 0\} \cup \{0\}\) is not.
Let \(A\) be a closed discrete
set inside a compact set \(K\). Show
that \(A\) is finite.
[Hint: assume for a contradiction that \(A\) is infinite.]
Use part (ii) to explain that one needs the “closed" hypothesis in part (iii).
Show that subset of a discrete metric space is discrete (hence the name).
Connected Sets
We call a metric space \(X\)
connected if the only subsets which are simultaneously
open and closed (that is “clopen") are \(X\) and the empty set \(\emptyset\). [Note that this is the
opposite situation to a discrete metric space, where subset is
clopen.]
Let \(X\) be the union of the intervals \([0,1)\) and \([2,3]\) together with the metric restricted from the standard metric on \(\mathbb R^n\). Show that \(X\) is not connected. [Note, this explains the terminology.]
Show that \(\mathbb R^n\) is
connected (so no proper subset of \(\mathbb
R^n\) is simultaneously open and closed).
[Hint: Assume not, so by definition we could write \(\mathbb R^n = U \cup V\) with \(U\) and \(V\) both open and non-empty, say \(x \in U, y \in V\). Consider the line
segment \(\ell(t):=x+t(y-x)\) with
\(t \in [0,1]\) from \(x\) to \(y\) and the “crossing point" from \(U\) to \(V\). ]
Matrices as metric spaces
We endow \(M_n(\mathbb R)\) , the set
of \(n \times n\) real matrices, with
the norm arising from viewing \(M_n(\mathbb
R)\) as \(\mathbb R^{n^2}\) ; so
\(\|A\| = \sqrt{\sum_{i,j}
|x_{ij}|^2}\) for any matrix \(A\) with entries \(x_{ij}\in\mathbb R\).
Explain why the determinant is continuous as a map from \(M_n(\mathbb R)\) to \(\mathbb R\).
Show that \(GL_n(\mathbb R)\) is open in \(M_n(\mathbb R)\).
Show that \(SL_n(\mathbb R)\) is closed in \(M_n(\mathbb R)\), but not compact.
Recall that \(O(n)\), the
orthogonal group, consists of column vectors which give a orthonormal
basis for \(\mathbb R^n\). Use
Heine-Borel to show that \(O(n)\) is
compact.
[Hint: To show that \(O(n)\) is
closed, find (finitely many) continuous functions \(f_i:O(n)\to \mathbb R\) and closed sets
\(K_i\subset \mathbb R\) such that
\(O(n)=\bigcap_i
f_i^{-1}(K_i)\).]
A different definition of compactness
Show that \(x\) being a limit point of a subsequence of a sequence \(\{x_n\}_{n \in \mathbb N}\) in a metric space \(X\) is the same thing as \(B_r(x)\) containing infinitely many terms \(x_n\) of the sequence for any choice of \(r > 0\).
We call a metric space \(X\) compact if whenever \(\{ U_i : \, i \in I \}\) is a collection of open subsets \(U_i \subseteq X\) with \(X = \bigcup_{i \in I} U_i\), then there exists a finite subset \(J \subseteq I\) with \(X = \bigcup_{i \in J} U_i\). [We say “Any open cover admits a finite subcover”.] \(\,\) Show that if \(X\) is compact then \(X\) is compact.
Suppose \(X\) is compact. Show that given \(\epsilon > 0\) there exists a finite set of points \(x_1, x_2, \ldots, x_r \in X\) such that \(X = \bigcup_{i=1}^r B_{\epsilon}(x_i)\). [This set of points is called a “finite \(\epsilon\)-net”.]
Suppose \(X\) is compact and
\(\{ U_i : \, i \in I \}\) is a
collection of open subsets \(U_i \subseteq
X\) with \(X = \bigcup_{i \in I}
U_i\). Show that there exists \(\epsilon > 0\) such that for any point
\(x \in X\) there exists \(i \in I\) with \(B_\epsilon(x) \subseteq U_i\).
[Such an \(\epsilon\) is called a
“Lebesgue number” for the cover \(\{ U_i : \,
i \in I \}.\)]
Hence, show that if \(X\) is compact then \(X\) is compact.
Remark: As mentioned in lectures, our notion of a set being compact is more commonly referred to as the set being “sequentially compact”. As shown, these two notions coincide for subsets of metric spaces, but they do not in general.