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Let $X$ be a metrizable space. I'd like to prove that if $X$ is pseudocompact, then $X$ is compact (the converse is true by the Heine-Borel theorem).

Suppose $X$ was not compact. Since $X$ is metrizable, we have that $X$ is not sequentially compact. Hence, there exists a sequence $\{x_n\}_{n\in\mathbb{N}}$ in $X$ such that $\{x_n\}_{n\in\mathbb{N}}$ doesn't have any convergent subsequences in $X$. Let $S$ be the support of the sequence $\{x_n\}_{n\in\mathbb{N}}$ and let $g: S\to\mathbb{R}$ the function defined by $g(x_n):=n$. Now I'd like to prove that $S$ is discrete. Why? Because if I will prove that $S$ is discrete, I had that $S$ contains vacuously its accumulation points, then $S$ is closed. Now the function $g:S\to\mathbb{R}$ is continuous (since $S$ is discrete). Hence, I can apply the Tietze Extension Theorem that tells me that $g$ can be extended to a continuous function $G:X\to\mathbb{R}$. But this function is not bounded since the restriction $G_{|S}=g$ is not bounded: this one is a contradiction with the hypothesis of pseudocompactness of $X$.

Can anyone help me please?

PS: "$X$ is pseudocompact" stands for "each continuous function $F:X\to\mathbb{R}$ is bounded".

José Carlos Santos
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Grace53
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  • What is the support of the sequence? Is it ${x_n\mid n\in\Bbb N}$? – José Carlos Santos Jul 12 '20 at 15:13
  • @JoséCarlosSantos I think you want the closure of that. – user10354138 Jul 12 '20 at 15:14
  • @JoséCarlosSantos Yes! The support of ${x_n}_{n\in\mathbb{N}}$ is the image of the sequence: the sequence formally is a function $x: \mathbb{N}\to X$, the image of this function, that is ${y\in X \mid y=x_n\text{ for some }n\in\mathbb{N}}$, is called the support of the sequence. – Grace53 Jul 12 '20 at 15:21
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    You technically need to clarify why you can obtain a well-defined function $g$. For example, what if $x_{m}=x_{n}$ for $m\neq n$? – halrankard Jul 12 '20 at 16:06
  • @halrankard You're right! If I define $g(x_n):=\min{m\in\mathbb{N}\mid x_m=x_n}$, is $g$ now well-defined? The min exists for the well ordering principle, right? – Grace53 Jul 13 '20 at 21:53
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    It is well-defined--so now you need to justify why it's unbounded. – halrankard Jul 13 '20 at 21:56
  • I have an idea: let's suppose absurdly that $g$ is bounded; "g is bounded" implies that there exists $n\in\mathbb{N}$ such that $\Omega:={m\in\mathbb{N}\mid x_m=x_n}$ is infinite; hence ${x_k}{k\in\Omega}$ is a constant (and then convergent) subsequence of the sequence ${x_n}{n\in\mathbb{N}}$, contraddiction. What do you think? @halrankard – Grace53 Jul 13 '20 at 22:27
  • Looks good to me! – halrankard Jul 13 '20 at 22:47

1 Answers1

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Suppose that $S$ is not discrete. Then there is some $p\in X$ such that $p$ is an accumulation point of $S$. In particular, there is some $n_1\in\Bbb N$ such that $d(x_{n_1},p)<1$. Since there are infinitely many $n$'s such that $d(x_n,p)<\frac12$, there is a natural number $n_2>n_1$ such that $d(x_{n_2},p)<\frac12$. Since there are infinitely many $n$'s such that $d(x_n,p)<\frac13$, there is a natural number $n_3>n_2$ such that $d(x_{n_3},p)<\frac13$. And so on. So, the sequence $(x_{n_k})_{k\in\Bbb N}$ is a subsequence of $(x_n)_{n\in\Bbb N}$ and $\lim_{k\to\infty}x_{n_k}=p$.

José Carlos Santos
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  • If there is an accumulation point $p$ of $S$, this doesn't mean that $p\in S$, we still don't know that $S$ is closed. My idea is to prove that $S$ is closed by prooving before that $S$ is discrete and not the converse. – Grace53 Jul 12 '20 at 15:30
  • Another remark: if $p$ is an accumulation point of $S$, this means that $p$ is the limit of some sequence ${s_k}{k\in\mathbb{N}}$ of $S$ but nobody tells me that ${s_k}{k\in\mathbb{N}}$ is a subsequence of ${x_n}_{n\in\mathbb{N}}$ – Grace53 Jul 12 '20 at 15:33
  • I've edited my answer. I hope that everything is correct now. – José Carlos Santos Jul 12 '20 at 15:33
  • Concerning your second comment, if you dont see why is it that $p$ is the limit of some subsequence of $(x_n)_{n\in\Bbb N}$, I can add that to my answer. – José Carlos Santos Jul 12 '20 at 15:36
  • ${s_k}$ is of the form $s_k=x_{n(k)}$ but nothing guarantees to me that $n:\mathbb{N}\to\mathbb{N}, k\mapsto n(k)$ is an increasing function. – Grace53 Jul 12 '20 at 15:37
  • I've added a proof of the fact that $p$ is indeed the limit of a subsequence of the sequence $(x_n)_{n\in\Bbb N}$. – José Carlos Santos Jul 12 '20 at 15:41
  • I've just read your proof. A question: since $p$ is an accumulation point of $S$, we know that the set ${n\in\mathbb{N}\mid d(x_n, p)<1/2}$ is not empty, why we are sure that this set is infinite? – Grace53 Jul 12 '20 at 15:45
  • Suppose otherwise. Then there is some $N\in\Bbb N$ such that $n\geqslant N\implies d(x_n,p)\geqslant\frac12$. Since there are only finitely many $n$'s such that $d(x_n,p)$ is greater than $0$ and smaller than $\frac12$, there is some $r>0$ such that you never have $d(p,x_n)<r$, unless $x_n=p$. This is impossible, since, by the definition of accumulation point, the ball ${x\in X\mid d(p,x)<r}$ must have elements of $S$ distince from $p$. – José Carlos Santos Jul 12 '20 at 15:55
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    Now everything is clear, you have been very kind. Thank you very much for your time and your patience! – Grace53 Jul 12 '20 at 16:00