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George F Simmons, Topology and Modern Analysis pg.79 Problem 4

Let $X$ and $Y$ be metric spaces. Show that an into mapping $f:X \rightarrow Y$ is continuous $\iff$ $f^{-1}\left(G\right)$ is closed in $X$ whenever $G$ is closed in $Y$.

I can prove the problem for open sets, and I have been trying hard for closed. However, seems like I am stuck somewhere missing something obvious. Please don't answer directly, just give a small hint if possible.

EDIT: I am using the definition that $f^{-1}\left(G\right)$ exists only when $f$ is onto and if it is not then $f^{−1}\left(G\right)$ is a loose term for $f^{-1}\left(H\right)$ where $H$ is the range of $f$ in $G$.

  • What is the definition of "closed" you are working with? – Zev Chonoles Jul 24 '12 at 06:25
  • Also, see this post for an answer in the more general context of topological spaces. – Zev Chonoles Jul 24 '12 at 06:27
  • Definition of closed is that the a contains all its limit points where $a$ is a limit point of $X$ if for every open sphere $S_{\epsilon}\left(a\right)$ there exists an $x \in X$ such that $x \in S_{\epsilon}\left(a\right)$ –  Jul 24 '12 at 06:55
  • I just don't understand this confusion about $f^{-1}(G)$ (or $f^{-1}[G]$, as I prefer to write). For $f:X\to Y$ and $G\subseteq Y$ the definition of $f^{-1}[G]$ is simply ${x\in X:f(x)\in G}$. I don't see any problem with that. – Stefan Geschke Jul 24 '12 at 12:04
  • I was not taught that way, not using that or equivalent definition, I was taught as I wrote in that comment. While, I did learn the new definition, once in a while the old habit unconsciously takes over. I hope, I get used to the new one fast enough though :-) –  Jul 25 '12 at 07:16

2 Answers2

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So you can show that $f$ is continuous iff preimages of open sets are open. Now you go from there. Closed sets are complements of open sets. What is $f^{-1}[Y\setminus G]$?

Stefan Geschke
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  • But the mapping is an into mapping, how do I know $f^{-1}$ exists for all Y –  Jul 24 '12 at 06:52
  • It's not relevant. You can always define $f^{-1} Y$ for any subset $Y$ of the codomain of $f$. – Zhen Lin Jul 24 '12 at 06:55
  • @Jayesh: The notation $f^{-1}[Y\setminus G]$ doesn't imply that $f^{-1}$ exists for all of $Y$. It's just the set of all preimages of elements of $Y\setminus G$, not all of which necessarily have preimages. Since all of $X$ is mapped into $Y$, we have that $f^{-1}[Y]$ is all of $X$, even though $f(X)$ need not be all of $Y$. – joriki Jul 24 '12 at 06:55
  • So lets say, $Z \in Y$ is the "range" of $f$ so that $f:X \rightarrow Z$ is an onto mapping. Then, if $G$ is closed in $Y$, then $G^{-1}$ is open in $Y$, but we are now considering $H = G^{-1} \bigcap Z$, how do I know that is also open? As far as I know, $H$ is open as a subset of $Z$ (which should be a subspace of $X$) in and only if it is the intersection with $Z$ of an open set which it is, but what about $H$ in $X$? –  Jul 24 '12 at 07:06
  • I guess, I am not using the definition of $f^{-1}\left(G\right)$ properly. I am using the definition that $f^{-1}\left(G\right)$ exists only when $f$ is onto and if it is not then $f^{-1}\left(G\right)$ is a loose term for $f^{-1}\left(H\right)$ where $H$ is the range. If someone one can make it more clear to me, it would be better. –  Jul 24 '12 at 07:13
  • I presume that by $G^{-1}$ you mean the complement of $G$? That's a very unusual notation, and particularly when we're talking about $f^{-1}$ it's rather confusing. The complement can be expressed as $Y\setminus G$ or $Y-G$, or, if you don't want to mention the underlying set with respect to which the complement is taken, $\overline G$. – joriki Jul 24 '12 at 07:15
  • I am very sorry, I meant to use $G'$, will correct it if possible. –  Jul 24 '12 at 07:17
  • Regarding $f^{-1}(G)$, as I wrote above, this is defined as the set of all preimages of elements in $G$, that is, all elements of $X$ whose images lie in $G$. It doesn't matter whether all elements of $G$ have preimages; some may have none, some may have one, some may have many. To distinguish this from the notation $f^{-1}(y)$ for a single element $y$, which is usually used to refer to the unique preimage of $y$, which only makes sense if $f$ is invertible, the set-wise preimage is often written as $f^{-1}[G]$, as in Stefan's answer. – joriki Jul 24 '12 at 07:18
  • Okay, then it is trivial. You just take the complements and get done. Somehow, now, the question is something else, while the actual difficulty was something else. I guess, I would edit the question and someone can edit his answer and then, I would accept it so that it represents what actually happened? –  Jul 24 '12 at 07:22
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The following steps lead to a solution:

(1) $G$ is closed in $Y$ iff $Y - G$ is open in $Y$.

(2) For any $A \subseteq Y$, we have $$f^{-1}(Y - A) = X - (f^{-1}(A)).$$

(3) Conclude that $f$ is continuous on $X$ $\iff$ $f^{-1}(G)$ is closed in $X$ whenever $G$ is closed in $Y$.

  • Your second step would be true only for onto mappings, while the textbook says $f$ is an into mapping, or am I missing something? –  Jul 24 '12 at 06:54
  • @Jayesh: See the comments under Stefan's answer. – joriki Jul 24 '12 at 06:56
  • @JayeshBadwaik You can define the preimage regardless of what kind of map $f$ is (injective, surjective or neither). (2) that I stated above always holds true. –  Jul 24 '12 at 07:16