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Let $(X,d)$ be a compact metric space. Let $f:X\to X$ be a function such that graph(f)=$(G(f))=\{(x,f(x):x\in X\}$ is closed.

Prove that $f$ is continuous.

TRY:

To show that $f$ is continuous enough to show that if $x_n\to x$ then $f(x_n)\to f(x)$.

Let $x_n\to x$ ;; since $X$ is compact so $f(x_n)$ has a convergent subsequence say $f(x_{n_k})\to y$. Consider the corresponding terms in the sequence $x_n$ then $(x_{n_k },f(x_{n_k}))\to (x,y).$

Now $(x_{n_k },f(x_{n_k}))\in G(f)$ which is closed hence $(x,y)\in G(f))$ which in turn implies that $y=f(x)$

So $f(x_{n_k})\to f(x)$.

But I need to show that $f(x_n)\to f(x)$

How can I show that??Please help me out.

Learnmore
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2 Answers2

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Suppose $x_n$ is a sequence converging to $x$, but $f(x_n)$ doesn't converge to $f(x)$.

Suppose $f(x_n)$ is not convergent at all.

In a compact space, a non convergent sequence has at least two distinct limit points. Thus you find subsequences $x_{n'_k}$ and $x_{n''_k}$ such that $f(x_{n'_k})\to y$ and $f(x_{n''_k})\to z$, with $y\ne z$. Since the graph is closed, you have both $(x,y)$ and $(x,z)$ in $G(f)$, a contradiction.

Therefore $f(x_n)$ is convergent. By your argument it converges to $f(x)$, because a subsequence does.

egreg
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You have founded a function $f\colon X\to X$ satisfying the following condition: for every sequence $(x_n)_n$ in $X$ with $x_n\to x$ there exist a subsequence $(x_{n_k})_k$ such that $f(x_{n_k})\to f(x)$.

Suppose $f$ is not continuous at $x$. Then there exists $\varepsilon>0$ such that for all $n\in\mathbb N$ there exists $y_n\in B(x,\frac1n)$ with $f(y_n)\notin B(f(x),\varepsilon)$. Therefore, the sequence $(y_n)_n$ do not satisfy the given condition ($(y_n)_n$ converges to $x$ but do not exist a subsequence $(y_{n_k})_k$ such that $f(y_{n_k})\to f(x)$).

Hence, $f$ (with such condition) must be continuous.

Let me elaborate it a little bit more.

We say that $f$ is continuous at $x$ if for every $\varepsilon > 0$ there exists $\delta > 0 $ such that $d(f(y),f(x))<\varepsilon$ whenever $d(y,x)<\delta$, or equivalently: $f(y)\in B(f(x),\varepsilon)$ whenever $y\in B(x,\delta)$.

When it was supposed that $f$ is not continuous at $x$, we got some $\varepsilon > 0$ such that for every $\delta > 0$ there exist $y\in B(x,\delta)$ with $f(y)\notin B(f(x),\varepsilon)$. In that way, I have just taken $\delta = 1/n$.

Rodrigo Dias
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  • Your proof is false consider $(-1)^n$ is divergent but it has a convergent subsequence $x_N=1$ – Learnmore Dec 13 '17 at 13:44
  • @Find_X: What are your $f$ and $X$ in case of your example? –  Dec 13 '17 at 13:59
  • @Find_X The important fact about $(y_n)$ is that there exists $\varepsilon>0$ such that $d(f(y_n),f(x)) >\varepsilon$ for all $n\in\mathbb N$. Thus, $f(x)$ is not a limit point of $(f(y_n))_n$. – Rodrigo Dias Dec 13 '17 at 14:40
  • @rldias; Please note my problem; I have got that for every sequence $x_n$ converging to $x$ there exists a subsequence of $f(x_n)$ namely $f(x_{n_k})$ converging to $f(x)$ – Learnmore Dec 13 '17 at 14:42
  • How to show that $f$ is continuous from here? – Learnmore Dec 13 '17 at 14:43
  • To show it ;let $x_n\to x$ such that $f(x_n)$ does not converge to $f(x)$ ,Now how to derive a contradiction from here? – Learnmore Dec 13 '17 at 14:44
  • By the hypothesis $f(x_n)$ has a subsequence converging to $f(x)$ .What next to ensure that $f(x_n)\to f(x)$ – Learnmore Dec 13 '17 at 14:45
  • Did you understand my doubt? – Learnmore Dec 13 '17 at 14:45
  • I've already discussed your doubt in my answer.. It was suppose that $f$ is not continuous at $x$ and it was built a sequence which is not compatible with this condition that $f$ satisfies. Can you see it? – Rodrigo Dias Dec 13 '17 at 14:56
  • I've edited my answer. Hope it is more helpful now. – Rodrigo Dias Dec 13 '17 at 15:10