Show the following statement from class: Let \(\left( X,d\right)\) be a metric space and let \(A\subseteq X\). Then \(A\) is closed if and only if any sequence of elements of \(A\) that converges has its limit in \(A\). In other words if \(x_n\in A\) for all \(n\in\mathbb N\) and \(\lim_{n\to\infty}x_n=x\) then \(x\in A\).
Let’s start by assuming that \(A\)
is closed and \(\left\lbrace
x_n\right\rbrace_{n\in\mathbb N}\) is a sequence of elements in
\(A\) that converges to an element
\(x\in X\). If \(x\not\in A\) then \(x\in A^c\). Since \(A\) is closed \(A^c\) is open and since \(x\in A^c\) we can find \(\epsilon_0>0\) such that \(B_{\epsilon_0}\left( x\right)\subseteq
A^c\). n other words \(B_{\epsilon_0}\left( x\right)\cap
A=\emptyset\). This is impossible since we know that due to the
convergence of \(\left\lbrace
x_n\right\rbrace_{n\in\mathbb N}\) we can find \(N\in\mathbb N\) such that for all \(n>N\) we have that \(x_n \in B_{\epsilon_0}\left( x\right)\) but
also \(x_n\in A\).
Conversely, let us assume that any limit of any sequence of elements in
\(A\) remains in \(A\) and show that \(A\) is closed. It is sufficient to show
that \(A^c\) is open. Consider \(x\in A^c\) and assume that there is no
\(\epsilon>0\) such that \(B_{\epsilon}(x)\subseteq A^c\). In other
words, for any \(\epsilon>0\) we can
find \(x_\epsilon\in A\) such that
\(x_\epsilon\in B_{\epsilon}\left(
x\right)\). This means that for any \(n\in\mathbb N\) we have \(x_n\in A\) such that \(x_n\in B_{1/n}(x)\). By definition this
means that \(d\left( x_n,x\right)<
\frac{1}{n}\) which implies that \(\left\lbrace x_n\right\rbrace_{n\in\mathbb
N}\) converges to \(x\). Since
\(\left\lbrace x_n\right\rbrace_{n\in\mathbb
N}\) is a sequence of elements from \(A\) we conclude by our assumption that
\(x\in A\) which is a
contradiction.
(Assignment sheet 2 problem 15) Show that if a sequence \(\{x_n\}\) converges in a discrete metric space, then it is eventually constant.
Find a metric space \(\left( X,d\right)\) and a set \(A\subseteq X\) such that \(A\) is closed and bounded but is not sequentially compact.
(i) Assuming that \(x_n \to x_0\),
we conclude that there exists \(N\in\mathbb
N\) such that for all \(n>N\)
we have that \(x_n\in
B_{\frac{1}{2}}(x)\). We saw in the discrete metric \(B_{\frac{1}{2}}(x)=\left\lbrace
x\right\rbrace\) and as such for all \(n>N\) we have that \(x_n=x\) which shows the desired
result.
(ii) Consider the set \(X=[0,1]\) we
the discrete metric. \(X\) is bounded
since \(X\subseteq B_2(0)\) and is
closed since it is the entire space (in fact every set is closed in the
discrete metric). Consider the sequence \(x_n=\frac{1}{n}\). The sequence is in \(X\) but since \(x_n\not=x_m\) for any \(n\not=m\) we see that it has no subsequence
that is eventually constant. Consequently it has no subsequence that
converges in the discrete metric so the set is not sequentially
compact.
(Assignment sheet 4 problem 4) The real axis and the imaginary axis divide \(\mathbb C\) into four quadrants as follows:
c|c \(\Omega_2\) & \(\Omega_1\) \(\Omega_3\) & \(\Omega_4\)
Determine the images of \(\Omega_1, \Omega_2, \Omega_3, \Omega_4\) under the exponential map \(z\mapsto e^z\).
We have that \(e^z = e^x e^{iy}\) so we know that segment of length more than \(2\pi\) of the line \(x=c\) goes into a circle of radius \(e^c\). Consequently \(e^z\) takes \(\Omega_1\) and \(\Omega_4\) to the set \[\left\lbrace w\in \mathbb C\;:\; e^0<{\left|w\right|}<e^\infty\right\rbrace = \left\lbrace w\in \mathbb C\;:\; {\left|w\right|}>1\right\rbrace\] and it takes \(\Omega_2\) and \(\Omega_3\) to the set \[\left\lbrace w\in \mathbb C\;:\; 0<{\left|w\right|}<e^0\right\rbrace = \left\lbrace w\in \mathbb C\;:\; 0<{\left|w\right|}<1\right\rbrace.\]
(Assignment sheet 4 problem 6(b)) Give examples to illustrate that, in general, for complex numbers \(z\), \(w\), \[Log (zw)\ne Log (z)+Log (w).\]
By definition \[Log(\xi) = \log\left( {\left|\xi\right|}\right)+ iArg\left( \xi\right).\] As such \[Log\left( zw\right) = \log\left( {\left|zw\right|}\right)+iArg\left( zw\right) = \log\left( {\left|z\right|}\right)+\log\left( {\left|w\right|}\right)+iArg\left( zw\right)\] and \[Log (z)+Log (w) = \log\left( {\left|z\right|}\right)+iArg(z)+ \log\left( {\left|w\right|}\right)+iArg(w).\] This means that \[Log (zw)\ne Log (z)+Log (w)\] if and only if \[Arg\left( zw\right)=Arg(z)+Arg(w)\] which we know is not true. For instance, for \(z=w=e^{-i\frac{\pi}{2}}= -i\) we have that \[Arg(z)=Arg(w) = -\frac{\pi}{2}\] but \[Arg\left( zw\right) = Arg(-1) = \pi.\]