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Let $f:X\rightarrow Y.$If $f $ is continuous,then for every convergent sequence $x_n\rightarrow x$ in $X$,the sequence $f(x_n)$ converges to $f(x)$.The converse holds if $X$ is first countabe.

I was getting problem in proving the converse part but Convergence of $f(x_n)$ implies convergence of $x_n$ (From Daniel Fischer's 2nd comment) cleared my query upto very very extent.Now i wanted to know where does the notion of first countability is used in proving its converse?

This is the proof given in Munkres

To prove the converse,assume that the convergent sequence condition is satisfied.Let A be as subset of $X$;We show that $f(\bar{A}) \subset \overline{f(A)}$.If $x\in \bar{A},$there is a sequence converging to $x$.By assumption,the sequence $f(x_n)\rightarrow f(x) $.Since $f(x_n)\in f(A) \implies f(x)\in \overline{f(A)}$.Hence $f(\bar{A}) \subset \overline{f(A)}$.

How does $f(\bar{A}) \subset \overline{f(A)}$ happened?

My trial:

Let $x\in \bar{A} $ then $f(x)\in f(\bar{A})-----------------(1)$ .

Now we'll show that $f(x)\in \overline{f(A)} $. Since $x\in \bar{A}$ so there exists a sequence,$<x_n>$ in $A$ such that $<x_n>\rightarrow x$ and by assumption we have $f(x_n)\rightarrow f(x)$ then $f(x)\in \overline{f(A)}---------------------------(2)$

From (1) and (2) we have $f(\bar{A}) \subset \overline{f(A)}$.

Is it correct?

Picaso
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  • I don't understand your question. Are you confused how this proof shows that $f(\bar A)\subset \overline{f(A)}$ or are you confused why showing this proves continuity? – TY Mathers Apr 30 '17 at 11:24
  • @SheafKeef:I'm confuse in how this proof shows $f(\bar{A}) \subset \overline{f(A)}$. – Picaso Apr 30 '17 at 11:33
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    Note that $f$ is continuous if and only if $f(\overline{A}) \subset \overline{f(A)}$ for any subset $A \subset X$. – Victor Chen Apr 30 '17 at 11:48

2 Answers2

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We want to show that $f(\bar A)\subset\overline{f(A)}$. That is, we want to show that if $x\in\bar A$ then $f(x)\in\overline{f(A)}$.

What's the definition of saying $x\in\bar A$? This is where we need first countability. Under the assumption of first countability, it's a fact that $x\in\bar A$ if and only if there is a sequence of points $x_n\in A$ such that $x_n\to x$.

Now, to show $f(x)\in\overline{f(A)}$, we need to show there is a sequence of points $y_n\in f(A)$ such that $y_n\to f(x)$. But using the assumption of the problem, since $x_n\to x$ we have $f(x_n)\to f(x)$, and we can then take $y_n=f(x_n)\in f(A)$, showing $f(x)\in\overline{f(A)}$, as desired.

TY Mathers
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  • :Let $x\in \bar{A} $ then $f(x)\in f(\bar{A})$ .

    Now we'll show that $f(x)\in \overline{f(A)} $. Since $x\in \bar{A}$ so there exists a sequence,$<x_n>$ in $A$ such that $<x_n>\rightarrow x$ and by assumption we have $f(x_n)\rightarrow f(x)$ then $f(x)\in \overline{f(A)}$

    From (1) and (2) we have $f(\bar{A}) \subset \overline{f(A)}$.

    Is it correct?

     – Picaso Apr 30 '17 at 12:50
  • @Picaso Yes that's exactly it. – TY Mathers Apr 30 '17 at 12:52
  • :I'm not getting the second last line of your answer:How does $x_n\rightarrow x \implies f(x_n)\rightarrow f(x)$ until $f$ is continuous? – Picaso Apr 30 '17 at 13:02
  • @Picaso that's the assumption!!! We're trying to show that if $X$ is first countable and $x_n\to x$ implies $f(x_n)\to f(x)$, then $f$ is continuous. – TY Mathers Apr 30 '17 at 13:07
  • @Sheef Keef:Sorry, I misread the statement.THanks a lot for the discussion. Without you this problem would have remained a disturbing problem. – Picaso Apr 30 '17 at 13:10
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Let it be that $f$ is not continuous at $x$.

Then some open set $U$ exists with $f(x)\in U$ and such that $f^{-1}(U)$ is not a neighborhood of $x$.

Now let $\mathcal B=\{B_1,B_2,\dots\}$ denote a countable neighborhood basis of $x$ with $B_1\supset B_2\supset\cdots$.

$B_n\setminus f^{-1}(U)\neq\varnothing$ for every $n$ and picking some $x_n\in B_n\setminus f^{-1}(U)$ gives a sequence $(x_n)_n$ with $x_n\to x$ together with $f(x_n)\notin U$ for each $n$. Then we will not have $f(x_n)\to f(x)$.

The fact that $x$ has a countable neighborhood base makes it possible to construct such a sequence under the condition that $f$ is not continuous at $x$.

drhab
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