Complex Analysis II, Michaelmas 2024. Problem Class 1

From its definition we have that \(d\left( x,y\right)\geq 0\) and \(d(x,y)=0\) if and only if \(x=y\). Moreover \[d\left( x,y\right) = \begin{cases} 0,& x=y,\\ 1,& x\not=y, \end{cases}=\begin{cases} 0,& y=x,\\ 1,& y\not=x, \end{cases}=d\left( y,x\right).\] We are left with showing the triangle inequality. Let \(x,y\) and \(z\) be given. We have that \[d\left( x,z\right)+d\left( z,y\right) \;=\; \begin{cases} 0, & x=y\text{ and }z=y,\\ 2, & x\not=z\text{ and }z\not=y,\\ 1, & \text{otherwise}, \end{cases}\;=\;\begin{cases} 0, & x=y=z,\\ \geq 1, & \text{otherwise}, \end{cases} \geq d\left( x,y\right)\] which shows the first part of the problem.
To see that the distance is not induced by a norm we notice that if it was \[d\left( x,0\right) = \left\lVert x\right\rVert\] and consequently for any scalar \(\lambda\) \[d\left( \lambda x,0\right) = {\left|\lambda \right|}\left\lVert x\right\rVert.\] If \(x\not=0\) then for any \(\lambda \not=0\) we have that \(\lambda x\not=0\) and as such \(d\left( \lambda x,0\right)=1\). This implies that for any \(x\not=0\) and \(\lambda\not=0\) we have that \[1={\left|\lambda\right|}\left\lVert x\right\rVert\] which is impossible. Hence, the distance is not induced by a norm.

We start by noticing that \(|f(t)-g(t)| \geq 0\) and as such \(\int_a^b|f(t)-g(t)| \, dt \geq \int_a^b0 \, dt = 0\). Moreover, since both \(f\) and \(g\) are continuous, so if \({\left|f-g\right|}\) and \[0=d_1\left( f,g\right) = \int_a^b|f(t)-g(t)| \, dt.\] implies, due to the continuity and non-negativity, that \(|f(t)-g(t)|=0\) for all \(t\in [a,b]\) or that \(f\equiv g\).
Since \({\left|f-g\right|}={\left|g-f\right|}\) we have that \(d_1\left( f,g\right)=d_1\left( g,f\right)\) and since for any \(f,g\) and \(h\) we have that \[{\left|f(t)-g(t)\right|} = {\left|\left( f(t)-h(t)\right)+\left( h(t)-g(t)\right)\right|} \leq {\left|f(t)-h(t)\right|} +{\left|h(t)-g(t)\right|}\] we conclude that \[d_1\left( f,g\right)\leq d_1\left( f,h\right)+d_1\left( h,g\right)\] which is the desired triangle inequality.

(i) We need to show that the complement of \(\{x\}\) is open. Take \(y\in \left\lbrace x\right\rbrace^c\), i.e. \(y\not=x\), and let \(\epsilon = \frac{d\left( x,y\right)}{2}\). We claim that \(B_\epsilon(y)\subseteq \left\lbrace x\right\rbrace^c\). Indeed, if \(x\in B_{\epsilon}(y)\) then \[d\left( x,y\right) < \epsilon =\frac{d\left( x,y\right)}{2}\] which is impossible.

(ii) We need to show that complement of \(\overline{B}_r(x)\) is open. By definition \[\overline{B}_r(x)^c=\left\lbrace y\in X\;|\; d\left( x,y\right)\leq r\right\rbrace^c = \left\lbrace y\in X\;|\; d\left( x,y\right)> r\right\rbrace.\] Let \(y\in \overline{B}_{r}(x)^c\). We know that \(d\left( x,y\right)>r\) and can define \[\epsilon=\frac{d\left( x,y\right)-r}{2}>0.\] We claim that \(B_{\epsilon}\left( y\right)\subseteq \overline{B}_r(x)^c\). Indeed, assume that \(B_\epsilon\left( y\right)\cap \overline{B}_r\left( x\right)\not=\emptyset\). If \(z\in B_\epsilon\left( y\right)\cap \overline{B}_r\left( x\right)\) then \[d\left( x,y\right) \leq d\left( x,z\right)+d\left( y,z\right) \leq r+d\left( y,z\right)< r + \epsilon = \frac{r+d\left( x,y\right)}{2}.\] This implies that \(d\left( x,y\right)< r\) which is impossible.
Alternative proof: By the triangle inequality we know that for any \(x,y\) and \(z\) we have that \[d\left( x,z\right) \leq d\left( x,y\right)+d\left( y,z\right)\] and \[d\left( y,z\right) \leq d\left( x,y\right)+d\left( x,z\right).\] Combining the two inequalities we find that \[d\left( x,y\right) \geq \max\left\lbrace d\left( x,z\right)-d\left( y,z\right),d\left( y,z\right)-d\left( x,z\right)\right\rbrace={\left|d\left( x,z\right)-d\left( y,z\right)\right|}.\] Thus, with \(\epsilon=\frac{d\left( x,y\right)-r}{2}\), we see that if \(z\in B_{\epsilon}(y)\) \[d\left( x,z\right) \geq d\left( x,y\right)-d\left( y,z\right)>d\left( x,y\right)-\epsilon=\frac{r+d\left( x,y\right)}{2}>r\] showing that \(B_\epsilon(y)\subseteq \overline{B}_r(x)^c\).

Consider any set \(X\) that is not a singleton with the discrete metric and let \(x\in X\) be arbitrary. We notice that \[\overline{B}_r\left( x\right) = \left\lbrace y\in X\;:\; d\left( x,y\right)\leq r\right\rbrace = \begin{cases} \left\lbrace x\right\rbrace,& r<1,\\ X, & r\geq 1 \\ \end{cases}\] Thus \(B_1\left( x\right) = X\). On the other hand, as we saw in class \[B_r\left( x\right) = \left\lbrace y\in X\;:\; d\left( x,y\right)< r\right\rbrace = \begin{cases} \left\lbrace x\right\rbrace,& r\leq 1,\\ X, & r> 1 \\ \end{cases}\] which implies that \(B_1\left( x\right)=\left\lbrace x\right\rbrace\). We claim that \(\overline{B_1\left( x\right)}=\left\lbrace x\right\rbrace\not=X\) which will give us the desired example. We have several way to show this but we will go with the original definition: \(\overline{A} = \left( \left( A^c\right)^0\right)^c\). Recall that with the discrete metric every set is open. We claim that for any open set \(A\) we have that \(A=A^0\). Once this is proven we’ll find that since every set is open in the discrete metric, we have that \[\overline{A} = \left( \left( A^c\right)^0\right)^c=\left( A^c\right)^c=A\] which shows the desired result.
Proof of the claim By definition, \(A^0\subseteq A\) so we only need to show that \(A\subseteq A^0\). Let \(x\in A\). Since \(A\) is open we can find an open ball of radius \(\epsilon\) for some \(\epsilon>0\) that is centred in \(x\) and is contained in \(A\). In other words, there exists \(\epsilon>0\) such that \(B_\epsilon(x)\subseteq A\). By the definition of \(A^0\) and the fact that pen balls are open sets we conclude that \(x\in A^0\). As \(x\) was arbitrary we find that \(A\subseteq A^0\) and the proof is complete.
Remark: If we use the theorem that states that \(A\) is closed if and only if \(A=\overline{A}\) we can conclude that any set is in the discrete metric satisfies \(A=\overline{A}\) since every set is closed.