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The graph of $f$ is $G(f) = \{(x,f(x)) : x\in X\} \subseteq X\times Y$

$X$ and $Y$ are metric spaces.

a) Suppose $f$ is continuous and prove that $G(f)$ is a closed set.

b) Suppose that $G(f)$ is compact and prove that $f$ is continous

For a), the definition of a closed set that comes to my mind is a set that contains all its limit points (or was it accumulation points?), is there another equivalent definition that may b more helpful to prove a)? Is it possible to prove this directly? Because at first glance the only way I could imagine to prove this is by contradiction or contrapositive.

I imagine that the proof of b) will be immediately derived from a).

Anupam
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JohanLiebert
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    You tagged your question as "real-analysis". What are $X$ and $Y$? Are they any topological spaces? – Philippe Malot Oct 18 '13 at 17:46
  • they are metric spaces. Thanks for pointing that out. @girianshiido – JohanLiebert Oct 18 '13 at 17:51
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    For metric spaces, you can use the convenient characterisation of closed sets. A set is closed off it is sequentially closed, i.e. iff it contains the limits of every convergent sequences with terms in $S$. It's also easy to prove the continuity of $f$ using sequences. – Philippe Malot Oct 18 '13 at 17:55

3 Answers3

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a) Let $(z_n)=(x_n,f(x_n))$ be a convergent sequence of $G(f)$. If $(x,y)$ is its limit, show that $y=f(x)$.

b) Let $x\in X$ and $(x_n)$ a convergent sequence with limit $x$. You have to prove that $(f(x_n))$ is convergent in $Y$ with limit $f(x)$. Use the sequence $z_n=(x_n,f(x_n))$ and use the fact that $G(f)$ is compact to prove that $(f(x_n))$ has $f(x)$ as an accumulation point. Then prove that any subsequence of $(f(x_n))$ has $f(x)$ as an accumulation point.

Philippe Malot
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  • Can you expand on part a, please. How does showing that y = f(x) complete the proof? – JohanLiebert Oct 18 '13 at 21:08
  • It proves that $(x,y)=(x,f(x))\in G(f)$. So every convergent sequence $(z_n)\in G(f)^{\mathbb N}$ has its limit in $G(f)$. Hence $G(f)$ is closed. – Philippe Malot Oct 18 '13 at 21:12
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    Ok I see. Could you sketch the proof that the limit $(x,y) = (x,f(x))$? – JohanLiebert Oct 18 '13 at 21:34
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    If $(z_n)$ is convergent with limit $(x,y)$, then $(x_n)$ and $f(x_n)$ are both convergent with limits $x$ and $y$ respectively. Since $f$ is continuous, $y=\lim f(x_n)=f(x)$. – Philippe Malot Oct 18 '13 at 21:48
  • One small thing, how do we know that there exists a convergent sequence in $G(f)$? Is it just a given? – JohanLiebert Oct 18 '13 at 22:08
  • A subset $C$ of a metric space $X$ is closed iff all convergent sequences with terms in $C$ have their limits in $C$. So it's a given. – Philippe Malot Oct 18 '13 at 22:27
  • In part b, how does the compactness of $G(f)$ imply that $f(x)$ has is an accumulation point of $f(x_n)$? – JohanLiebert Oct 18 '13 at 22:36
  • let us continue this discussion in chat – Philippe Malot Oct 18 '13 at 22:48
  • I have a question for part b). There is subsequence $(x_{n_k},f(x_{n_k})) \to (x,y)\in G_f$. Then we will have $y=f(x)$ but how do I prove that $f(x_n) \to f(x)$? It is true that every subsequence of $f(x_n)$ has a subsequence converging to $f(x)$. Can you please elaborate that part? I have also posted the question here. https://math.stackexchange.com/questions/3967999/f-is-continuous-iff-gf-is-a-closed-set-in-metric-spaces – Ri-Li Dec 31 '20 at 13:15
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Hint: For (a), every metric space is Hausdorff, the result is frue for any Hausdorff space $Y$. Choose any (x,y)\in $X\times Y\setminus G(f)$. Then $x\in X$ and $y\ne f(x)$. Use Hausdorff condition in $Y$.

Anupam
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We want to show that every sequence which converges in $X\times Y$ has a limit in $G(f)$.

A sequence $(x_n, y_n)$ converges to $(x, y)$ if and only if $x_n\rightarrow x$ and $y_n\rightarrow y$.

Note that $(x_n, y_n)\in N_\epsilon(x, y)\implies(x_n)\in N_\epsilon(x) $ and $(y_n)\in N_\epsilon(y)$ respectively.

Conversely, if $x_n \in N_{\epsilon/2}(x)$ and $y_n \in N_{\epsilon/2}(y)$, then $(x_n, y_n)\in N_\epsilon(x, y)$.

So now we see that if $(x_n, y_n)\in G(f)$, $(x_n,y_n) \rightarrow (x,y)$, then $y_n\rightarrow f(x_n)$ as defined by $G(f)$ and $x_n \rightarrow x, f(x_n)\rightarrow y.$

Since $f$ is assumed to be continuous, $f(x_n)\rightarrow f(x)$ so $y=f(x)$. Therefore $(x,y)\in G(f)$ and we conclude $G(f)$ is closed.

YoTengoUnLCD
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grayQuant
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