Exercise 18.9

Proof. We show using induction that $f$ is continuous for any collection ${\left\{A_{\alpha }\right\}}_{\alpha =1}^n$, for any $n\in {\mathbb{Z}}_{\text{+}}$, where each $A_{\alpha }$ is closed. This of course shows the desired result since the collection is ${\left\{A_{\alpha }\right\}}_{\alpha =1}^n$ for some $n\in {\mathbb{Z}}_{\text{+}}$ if it is finite. So first, for $n=1$, we have that $A_1={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha =1}^n{A}_{\alpha }=X$ so that of course $f=f\upharpoonright X=f\upharpoonright A_1$ is continuous.

Now suppose that $f$ is continuous for any collection of size $n$ and suppose we have the collection ${\left\{A_{\alpha }\right\}}_{\alpha =1}^{n+1}$ of size $n+1$. Let $A={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha =1}^n{A}_{\alpha }$, which is closed by Theorem 17.1 since each $A_{\alpha }$ is closed and it is a finite union, and let $B=A_{n+1}$ so that $B$ is also closed. We then have that $A\cup B={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha =1}^n{A}_{\alpha }\cup A_{n+1}={\bigcup }_{\alpha =1}^{n+1}{A}_{\alpha }=X$. We know that $g=f\upharpoonright B=f\upharpoonright A_{n+1}$ is continuous. Considering the set $A$ as a subspace of $X$, then each $A_{\alpha }$ for $\alpha \in \left\{1,\dots ,n\right\}$ is closed in $A$ by Theorem 17.2 since they are subsets of $A$ and closed in $X$. Since by definition ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha =1}^n{A}_{\alpha }=A$, it follows from the induction hypothesis that $f^{\prime }=f\upharpoonright A$ is continuous. Clearly also for any $x\in A\cap B$ we have that $x\in A$ and $x\in B=A_{n+1}$ so that $f^{\prime }(x)=(f\upharpoonright A)(x)=f(x)=(f\upharpoonright A_{n+1})(x)=g(x)$.

Then by the pasting lemma the function $h:X\to Y$ is defined by

is continuous. However, consider any $x\in X$. If $x\in A$ then $h(x)=f^{\prime }(x)=(f\upharpoonright A)(x)=f(x)$. Similarly if $x\in B$ then $h(x)=g(x)=(f\upharpoonright B)(x)=f(x)$ as well. This suffices to show that $h=f$ since $x$ was arbitrary. Thus $f$ is continuous, which completes the induction. □

Proof. First, it is trivial to show that the collection covers all of $ℝ$, i.e. that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{A}_n=ℝ$. It is also obvious by this point that each $A_n$ is closed in the standard topology. Clearly, $f$ is not a continuous function since there is a discontinuity at $x=0$, which is trivial to prove. Lastly, consider any $n\in {\mathbb{Z}}_{\text{+}}$. If $n=1$ then for any $x\in A_n=A_1=(-\infty ,0]$ we have that $x\le 0$ and hence $f(x)=1$. Likewise if $n=2$ then for any $x\in A_n=A_2=[1,\infty )$ it follows that $x\ge 1>0$ , and hence $f(x)=0$. Lastly, if $n>2$ then for any $x\in A_n=[1∕(n-1),1∕(n-2)]$ we have that $0<1∕(n-1)\le x$ so that $f(x)=0$ again. Thus in all cases $f\upharpoonright A_n$ is constant and therefore continuous. This shows the desired properties. □

Proof. Consider any $x\in X$ so that there is a neighborhood $U^{\prime }$ of $x$ that intersects a finite subcollection ${\left\{A_k\right\}}_{k=1}^n$ of the full collection $\left\{A_{\alpha }\right\}$. Consider $A={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^n{A}_k$ as a subspace of $X$, from which it follows from Theorem 17.2 that each $A_k$ is closed in $A$ since it is a subset of $A$ and closed in $X$. It is then easy to show that $U^{\prime }\subset A$. It also follows from part (a) that $f\upharpoonright A$ is continuous with the domain being the subspace topology on $A$.

Now consider any neighborhood $V$ of $f(x)$, noting that of course $x\in A$ since $x\in U^{\prime }$ and $U^{\prime }\subset A$. Thus $f(x)$ is in the image of $f\upharpoonright A$ so that there is a neighborhood $U_A$ of $x$ in the subspace topology such that $(f\upharpoonright A)(U_A)\subset V$ by Theorem 18.1 since $f\upharpoonright A$ is continuous. Since $U_A$ is open in the subspace topology, there is an open set $U_X$ in $X$ where $U_A=A\cap U_X$. Now let $U=U^{\prime }\cap U_X$, which is open in $X$ since $U^{\prime }$ and $U_X$ are both open in $X$. Then also $x\in U_X$ since $x\in U_A$ and $U_A=A\cap U_X$, and hence $x\in U$ since also $x\in U^{\prime }$ and $U=U^{\prime }\cap U_X$. Thus $U$ is a neighborhood of $x$ in $X$.

Let $z$ be any element of $f(U)$ so that $z=f(y)$ for some $y\in U$. Then $y\in U^{\prime }$ and $y\in U_X$ since $U=U^{\prime }\cap U_X$. Then also $y\in A$ since $U^{\prime }\subset A$, and hence $y\in A\cap U_X=U_A$. From this it follows that $z=f(y)=(f\upharpoonright A)(y)\in (f\upharpoonright A)(U_A)$ so that $z\in V$ since $(f\upharpoonright A)(U_A)\subset V$. Since $z$ was arbitrary, this shows that $f(U)\subset V$, which in turn shows that $f$ is continuous by Theorem 18.1 since $V$ was an arbitrary neighborhood of $f(x)$ and $x$ was an arbitrary element of $X$. □