Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $G$.

share|improve this question
    
What is the definition of "closed" you are working with? –  Zev Chonoles Jul 24 '12 at 6:25
    
Also, see this post for an answer in the more general context of topological spaces. –  Zev Chonoles Jul 24 '12 at 6:27
    
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 $x \in S_{\epsilon}\left(a\right)$ –  Jayesh Badwaik Jul 24 '12 at 6:55
    
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 :-) –  Jayesh Badwaik Jul 25 '12 at 7:16
add comment

2 Answers

up vote 2 down vote accepted

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 $f^{-1}[Y\setminus G]$?

share|improve this answer
    
But the mapping is an into mapping, how do I know $f^{-1}$ exists for all Y –  Jayesh Badwaik Jul 24 '12 at 6:52
    
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 6:55
    
@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 6:55
    
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 $X$? –  Jayesh Badwaik Jul 24 '12 at 7:06
    
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. –  Jayesh Badwaik Jul 24 '12 at 7:13
show 4 more comments

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 $Y$.

share|improve this answer
    
Your second step would be true only for onto mappings, while the textbook says $f$ is an into mapping, or am I missing something? –  Jayesh Badwaik Jul 24 '12 at 6:54
    
@Jayesh: See the comments under Stefan's answer. –  joriki Jul 24 '12 at 6:56
    
@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. –  fpqc Jul 24 '12 at 7:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.