Let $f: (X,d) \rightarrow (Y, d')$ be a function. Prove that the following are equivalent:

  1. $f$ is continuous .

  2. For every $A \subset X$, $f(cl(A)) \subset cl(f(A))$.

  3. For every closed set $B$ in $Y$, the set $f^{-1}(B)$ is closed.

My proof:

"1 $\Rightarrow 2$":

Suppose $f$ is continuous at $a \in X$. Let $y \in f (cl(A)) \Rightarrow f^{-1}(y) \subset cl(A)$. Then there exists $ x_n \in A$: $x_n \Rightarrow f^{-1}(y)$. Let $a=f^{-1}(y)$. Since $f$ is continuous, $x_n \in A$ and $x_n \rightarrow a \Rightarrow f(x_n) \rightarrow f(a)$. Since $x_n \in A \Rightarrow f(x_n) \in f(A)$ and $f(a) =y$. Then there exists $f(x_n) \in f(A) : f(x_n) \rightarrow y \Rightarrow y \in clf(A) \Rightarrow f(cl(A)) \subset cl(f(A))$. $\blacksquare$

"2 $\Rightarrow$ 3" Let $A \in X$ such that $f^{-1}(B) = A$. We have $A \subset cl(A)$. Let $x \in cl(A) \Rightarrow f(x) \in f(cl(A))$. Since $f^{-1}{B} = A \Rightarrow B = f(A)$. Since $B$ is closed $\Rightarrow B = cl(B) = f(A) = cl(f(A))$. From (2), $f(cl(A)) \subset cl(f(A)) = cl(B)= B \Rightarrow f(x) \in B \Rightarrow x \in f^{-1}(B)= A \Rightarrow x \in A. $ Then $cl(A) \subset A \Rightarrow A $ is closed. $\blacksquare$

"3 $\Rightarrow$ 1" This can be prove by changing closed set to open set, then by using the definition of continuous to prove.

How was my first 2 proofs?

  • $\begingroup$ 1 $\Rightarrow$ 2 doesn't make sense. $f^{-1}(y)$ is not defined as an element of $X$! $\endgroup$
    – D_S
    Jan 26 '16 at 5:40
  • $\begingroup$ Also, there are so many ways to formulate the definition of continuity, which definition are you using? $\endgroup$
    – D_S
    Jan 26 '16 at 5:41
  • 1
    $\begingroup$ I was wondering about that $f^{-1}(y)$ part as well. Since from $1 \Rightarrow 2$, I'm not supposed to use the definition of continuity that the preimage of an open/closed set is open/closed so I used the definition that if $x_n \Rightarrow a$, then $f(x_n) \Rightarrow f(a)$. $\endgroup$
    – cooselunt
    Jan 26 '16 at 5:46
  • $\begingroup$ I see. Well, isn't $cl(A)$ the set of points in $X$ which are limits of a sequence in $A$? (not sure if that's your definition of closure) If $x \in X$ is the limit of a sequence $x_n \in A$, then by continuity, $f(x)$ is the limit of the sequence $f(x_n) \in f(A)$. That makes $f(x)$ a limit point of $f(A)$, i.e. $f(x)$ is in the closure of $f(A)$. $\endgroup$
    – D_S
    Jan 26 '16 at 5:54
  • $\begingroup$ Oh yeah. I think If I change from y to f(x), then it can avoid the mistake that this function might not be surjective? Then the rest should be the same proof. $\endgroup$
    – cooselunt
    Jan 26 '16 at 5:59

(a) If $U$ is open in $Y$, then $B=Y-U$ is closed So $f^{-1}(B)=X-f^{-1}(U)$ is closed iff $f^{-1}(U)$ is open.

Hence $(1)$ iff $(3)$.

(b) Define $cl(A):=\{ x\in X| d_X(x_n,x)\rightarrow x,\ x_n\in A \}$.

This is closed set : $y\in X-cl(A)$ Consider open $\frac{1}{n}$-ball $B:=B_\frac{1}{n} (y)$. If each $B$ contains $x_n\in cl(A)$, then $x_n\rightarrow y$. Since $x_n\in cl(A)$ there exists $d_X(x_n,z_n)< \frac{1}{n},\ z_n\in A$ That is $z_n\rightarrow y$. So $y\in cl(A)$. That is for some $n$, $B \cap cl(A)=\emptyset$ So we complete the proof.

So (3) implies that $f^{-1} ( cl(f(A)) )$ is closed And note that it contains $A$ In further it contains $cl(A)$ So $$ f(cl (A)) \subset cl (f(A) ) $$

That is $(3)$ implies $(2)$.

(c) $f(cl(A))\subset cl(f(A))$ iff $x_n\in A\rightarrow x$ implies that $f(z_n)\rightarrow f(x) $ and $z_n\in A$

That is $(2)$ implies the following : If $A=\{x_n\}$ where $x_n\rightarrow x$, then $f(x_n)$ has a convergent subsequence whose limit is $f(x)$.

We will show that $(2)$ implies $(3)$ : For closed $B$ in $Y$, assume that $f^{-1}(B)$ is not closed That is there exists $x_n$ s.t. $x_n=f^{-1}(y_n)\rightarrow x$ and $f(x)$ is not in $B$. By above, $y_n=f(x_n)$ has a convergent subsequence whose limit is $f(x)$. This is a contradiction.


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