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: &quot;$X$ is pseudocompact&quot; stands for &quot;each continuous function $F:X\to\mathbb{R}$ is bounded&quot;.
Let $X$ be a metric space. Prove that if every continuous function $f: X \rightarrow \mathbb{R}$ is bounded, then $X$ is compact.  This has been asked before, but all the answers I have seen prove the contrapositive. Realistically, this may be the way to go, but is there way to exhibit an unbounded continuous function (under the assumption $X$ is not compact) without appealing to results beyond introductory real analysis (e.g., the solution I've seen involves the Tietze Extension theorem)?  Because we're working with metric spaces, it's clear that the assumption that $X$ is noncompact (towards proving contrapositive) will lead us to extract a sequence of points in the space with no convergent subsequences. But how can we infer the existence of an unbounded continuous on the space knowing only about some sequence of points in this space (without using anything too advanced)?
If H is a subgroup of G ( | G:H |=n, n > 1), then how can you show that G has a factor group G/N which is isomorphic to a subgroup D of the symmetric group Sn with |D| > 1.  I've been looking at Cayley's Theorem but it's not specific to the factor groups. QT MySql connectivity using Windows XP, Qt Creator 4.5.2(windows 32 bit) Convert a router to a switch? Is it feasible to build a stream cipher from a cryptographic hash function? 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