Tutorial Note 7 (set operations with functions; topology on $\R$: openness, closedness and compactness)

Example 1.
(a) (i) For any $x\in (0,1)$, we can choose $\delta_x=\min\{x,1-x\}$ such that \[
(x-\delta_x,x+\delta_x)\subseteq [0,1],
\] therefore $x$ is an interior point of $[0,1]$.

(ii) For any $r\in (0,\infty)$, both $(-r,r)\not\subseteq [0,1]$ and $(1-r,1+r)\not \subseteq [0,1]$, so they are not interior point.

(b) By definition, the set $\R\setminus U$ is closed if and only if its complement $\R\setminus (\R\setminus U)=U$ is open.

(c) We note that \[
E+U=\cupp_{e\in E}(e+U).
\] But the translation of an open set is still open, it is because \[
e+U=e+\sqcupp (a_i,b_i)=\sqcupp (e+a_i,e+b_i),
\] is a union of (disjoint) open intervals. Therefore $E+U$ is open as it is a union of open sets.

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Example 2.

(a) It is closed since $\R\setminus \{0\}=(-\infty,0)\cup (0,\infty)$ is open.

(b) It is open since it is a union of open sets.

(c) It is closed since $\R\setminus \Z=\cupp_{n\in \Z}(n-1,n)$ is open.

(d) It is neither closed nor open.

If $\Q$ is open, pick $q\in\Q$, then there is $r>0$ s.t. $(q-r,q+r)\subseteq \Q$, but by density there is an $a\in \R\setminus \Q$, $a\in (q-r,q+r)\subseteq \Q$, a contradiction.

If $\Q$ is closed, then $\R\setminus \Q$ is open, again by density of $\Q$ this is impossible.

(e) It is closed since it is an intersection of closed sets $C_n$'s inductively defined below:

(f) It is not closed since \[
\R\setminus \left\{\frac{1}{n}:n\in \N\right\}=(-\infty,0]\cup \cupp_{n=1}^\infty \bigg(\frac{1}{n+1},\frac{1}{n}\bigg)\cup (1,\infty)
\] is not open since $0$ is not an interior point.

(g) It is closed since \[
\R\setminus \brac{\left\{\frac{1}{n}:n\in \N\right\}\cup \{0\}}=(-\infty,0)\cup \cupp_{n=1}^\infty \bigg(\frac{1}{n+1},\frac{1}{n}\bigg)\cup (1,\infty)
\] is a union of open sets. Note that we just subtract one more point in the set equality of part (f).

(h), (i) Both are open and closed by definition.

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Eaxmple 3.
Note that $x$ being a limit of some sequence in $A$ is that $x=\limn a_n$, for some $a_n\in A$.

(a) $\Rightarrow$ (b) Let $x$ be a limit of some sequence in $A$, i.e., $x=\limn a_n$, for some $a_n\in A$, we need to show $x\in A$. If not, i.e., $x\in \R\setminus A$, since $A$ is closed, $\R\setminus A$ is open, so there is an $r>0$ such that \[
(x-r,x+r)\subseteq\R\setminus A.
\] But $x=\limn a_n$, there must be an $N$ such that $n>N\implies |x-a_n|<r$, i.e., $a_n\in (x-r,x+r)$, this a a contradiction as there will be (infinitely many) $n$ such that $a_n\in (x-r,x+r)\subseteq \R\setminus A$ and $a_n\in A$.

(b) $\Leftarrow$ (a) To show $A$ is closed, we try to show $\R\setminus A$ is open. Let $y\in \R\setminus A$, we hope there is an $r>0$ such that $(y-r,y+r)\subseteq \R\setminus A$.

For the sake of contradiction let's suppose there is no such $r>0$, i.e., let's assume for each $r>0$, \[
(y-r,y+r)\not\subseteq \R\setminus A\iff (y-r,y+r)\cap A\neq \emptyset.\] In particular, for each $n$ we take $r=1/n$, then there will be an $a_n\in A$ such that \[a_n\in (y-1/n,y+1/n)\iff |y-a_n|<\frac{1}{n},\] therefore \[
y=\limn a_n.
\] By hypothesis $y\in A$, a contradiction to that $y\in \R\setminus A$ originally.

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Example 4.

$\bm{f(K)}$ is closed. We use the sequential criterion in Example 3. Indeed, let $x\in \R$ be s.t. $x=\limn f(k_n)$, for some $k_n\in K$. We try to show $x\in f(K)$. As $\{k_n\}$ is bounded, it has a convergent subsequence $\{k_{n_p}\}$ such that $k_{n_p}\to k$. Since $K$ is closed, $k\in K$ by Example 3. Therefore $x=\lim_{p\to\infty}f(k_{n_p})=f(k)$ by continuity, thus $x\in f(K)$.

$\bm{f(K)}$ is bounded. We apply Supremum Limit Theorem to $\sup f(K)$. If $\sup f(K)=\infty$, then there is a sequence in $x_n\in K$ such that $f(x_n)\to \infty$. But $\{x_n\}$ is bounded, it has a convergent subsequence $\{x_{n_k}\}$, $x_{n_k}\to k$, for some $k\in K$ (recall that since $K$ is closed, $k$ must be in $K$), therefore by continuity we have \[
f(k)=\lim_{k\to\infty}f(x_{n_k})=\infty,
\] a contradiction.

Similarly, we can show that $\inf f(K)>-\infty$.

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Example 5. Let $K=\capp_{n=1}^\infty K_n$. $K$ is of course compact since it is an intersection of closed sets and $K\subseteq K_1$. Now we show that $K\neq \emptyset$ by exhibiting an element in $K$.

For this, for each $n$ we pick $x_n\in K_n$,  as $\{x_n\}$ is bounded, it has a convergent subsequence $\{x_{n_k}\}$, $x_{n_k}\to x$. We show that $x\in K$. Indeed, for each fixed $p\in \N$, there is $N$ such that \[
k>N\implies n_k>p\implies K_{n_k}\subseteq K_p.
\] So for $k>N$, $x_{n_k}\in K_p$ and thus $x=\lim_{k\to\infty} x_{n_k}\in K_p$ by Example 3. Since $p$ is arbitrary, $x\in K:=\capp_{n=1}^\infty K_n$.

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Example 6. Fix $\epsilon>0$, for every $x\in [a,b]$, there is an $n\in \N$ such that \[
f(x)-f_n(x)<\epsilon,
\] thus $x\in \cupp_{n=1}^\infty (f-f_n)^{-1}(-\infty,\epsilon)$. Note that this is true for each $x\in [a,b]$, we have \[
[a,b]\subseteq \cupp_{n=1}^\infty (f-f_n)^{-1}(-\infty,\epsilon).
\]
Note that it becomes straightforward to create an open cover of $[a,b]$.

By hypothesis for each $n$ both $f$ and $f_n$ are continuous, so by Topological Continuity Theorem there is an open set $U_n\subseteq \R$ such that \[(f-f_n)^{-1}(-\infty,\epsilon)=U_n\cap [a,b].\] It follows that \[
[a,b]\subseteq \cupp_{n=1}^\infty U_n.
\] Since $[a,b]$ is compact, by Theorem 9 there is an $N\in \N$ such that \[
[a,b]\subseteq \cupp_{n=1}^N U_n.
\] Thus \[[a,b]=\cupp_{n=1}^N (U_n\cap [a,b])=\cupp_{n=1}^N (f-f_n)^{-1}(-\infty,\epsilon)=(f-f_N)^{-1}(-\infty,\epsilon).\] The last equality follow from the fact that $\big\{(f-f_n)^{-1}(-\infty,\epsilon)\big\}_{n=1}^\infty$ is ascending.

Now for every $x\in [a,b]$, for every $n\ge N$, \[
f(x)-f_n(x)\leq f(x)-f_N(x)<\epsilon,
\] thus $f_n\toto f$ on $[a,b]$.