If each $A_i$ is closed in $A$, show that a given function $f:A \to B$ is continuous, if and only if the restrictions $f_{A_i} \to B$ is continuous.

I'm not sure how to go about solving this question. I know that a set is only closed if its complement is open but I don't see how that helps me here unfortunately.

  • $\begingroup$ The difficulty of this question depends on how much mathematical background you have. Are you are allowed to use the standard fact that a function $f:A \to B$ is continuous if and only if for every subset $C$ closed in $B$, $f^{-1}(C)$ is closed in $A$? If so then it's fairly easy. You need also that a finite union of closed subsets of $A$ is closed in $A$. $\endgroup$ – Michael Cohen Feb 17 at 16:38
  • $\begingroup$ Also if you know the Pasting lemma or Gluing lemma that can be used to prove this easily. $\endgroup$ – absolute0 Feb 17 at 17:49
  • $\begingroup$ @MichaelCohen Would that mean that this wouldn't work if i was from 1 to infinity? It has to be finite? $\endgroup$ – user874917 Feb 18 at 0:41
  • $\begingroup$ @dk1233 Yes the index set has to be finite. However if you replaced the word "open" by "closed" throughout then the proposition would also work for an infinite index set. $\endgroup$ – Michael Cohen Feb 18 at 1:42
  • $\begingroup$ @MichaelCohen Can you explain why it wouldn't work for an infinite set? $\endgroup$ – user874917 Feb 18 at 2:23

If $f:A\longrightarrow B$ is continuous, obviously $f|_{A_i}$ is continuous for all $i\in\{1,\ldots,n\}$.

Now assume that all the $f|_{A_i}$, $i\in\{1,\ldots,n\}$ are continuous, and let $x_0\in A:=\bigcup_{i=1}^n A_i$. Let $i_1,\ldots,i_k\in\{1,\ldots,n\}$ be the indexes of all the sets $A_{i_j}$ such that $x_0\in A_{i_j}$.

Also, because the $A_i$ are all closed, $\mathbb{R}\setminus A_i$ are all open. Then, if $x_0\not\in A_i$ for $i\not\in\{i_1,\ldots,i_k\}$, we can take $\delta_i\in\mathbb{R}^+$ such that $B(x_0,\delta_i)\cap A_i=\emptyset$. Let then $\delta_M=\min\{\delta_i :i\not\in\{i_1,\ldots,i_k\}\}$. We have that $B(x_0,\delta)\cap A_i=\emptyset$ for all $i\not\in\{i_1,\ldots,i_k\}$.

Let $\epsilon\in\mathbb{R}^+$, and let $\delta_{i_j}\in\mathbb{R}^+$ such that $|f|_{A_{i_j}}(x)-f(x_0)|<\epsilon$ for all $x\in B(x_0,\delta_{i_j})\cap A_{i_j}$ and $j\in\{1,\ldots,k\}$.

If we choose $\delta:=\min\{\delta_M,\delta_{i_j} : j\in\{1,\ldots,k\}\}$, then, if $x\in B(x_0,\delta)\cap A$, $x\not\in A_i$ for all $i\not\in\{i_1,\ldots,i_j\}$ because $B(x_0,\delta)\cap A_i=\emptyset$. As $x\in A=\bigcup_{i=1}^n A_i$, there must exist $i_j\in\{i_1,\ldots,i_k\}$ such that $x\in A_{i_j}$. Then, $x\in B(x_0,\delta)\cap A_{i_j}\subset B(x_0,\delta_{i_j})\cap A_{i_j}$, and we get that: $$|f(x)-f(x_0)|=|f|_{A_{i_j}}(x)-f(x_0)|<\epsilon$$ Which concludes the proof.


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