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I came across something in the literature a few weeks ago (without proof):

If $X$ is a one-to-one continuous image of $[0,\infty)$, and $X$ is Hausdorff, then $X$ is metrizable.

How can you prove this?

EDIT: I originally left out the assumption that $X$ is a one-to-one image. You will need this!

Forever Mozart
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1 Answers1

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As far as I can see $X$ does not even have to be sequential.

Let $u$ be any point in $\beta [0,\infty) \setminus[0,\infty)$ and let $X$ be the quotient of the subspace $[0,\infty) \cup \{u\}$ identifying $u$ with $0$. Then $X$ is a Hausdorff space because $\{0, u\}$ is compact. The restriction of the quotient map to $[0,\infty)$ is a continuous bijection onto $X$.

The interval $[1, \infty)$ is not closed in $X$: since $[0,1]$ is compact, it is closed in $\beta[0,\infty)$, so $u$ must be a limit point of $(1,\infty)$.

However, $[1, \infty)$ is sequentially closed: it is closed in $[0,\infty)$, which, because it is normal, is sequentially closed in its Čech-Stone compactification. (proof)

Niels J. Diepeveen
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  • If I understand correctly, you have essentially given $[0,\infty)$ a coarser topology by enlarging the neighborhoods of $0$. Given a free ultrafilter $u$ on $\omega$, we can define a neighborhood of $0$ to be $[0,\epsilon)\cup \bigcup _{n\in A}(n,n+1)$ where $\epsilon>0$ and $A\in u$. The resulting space should be Hausdorff, but not first countable at $0$. Correct? – Forever Mozart Jun 22 '17 at 18:21
  • Or, given a sequence $s\in \omega^\omega$ we can define a neighborhood of $0$ to be $[0,\epsilon)\cup \bigcup _{n\in \omega}(s(n),s(n)+1)$ where $\epsilon>0$. – Forever Mozart Jun 22 '17 at 18:27
  • Explicitly constructing a neighbourhood base from a free ultrafilter on the positive integers was my first approach. Then I thought that I might as well use a free z-ultrafilter on $[0,1)$ a.k.a. a point in $\beta[0,1)$, which already carries a suitable neighbourhood base. The final result is very similar. – Niels J. Diepeveen Jun 22 '17 at 18:40
  • @ForeverMozart Seeing that this claim appears to be incorrect, may I ask what literature you found it in? Just curious. – M10687 Jun 22 '17 at 18:43
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    @M10687 It was an old paper (1960's maybe), I thought, but I haven't been able to find it again. Maybe I'm imagining things. – Forever Mozart Jun 22 '17 at 19:15