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Let X be Hausdorff space and X/~ be "~ Quotient Spaces",

Given that X/~ is T1,

Does it satisfy for X/~ to be Hausdorff?

Martin Sleziak
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Matan L
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1 Answers1

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Let $X$ be two copies of the real line, and let $X\to Y$ be the quotient map that identifies the two lines along the set $\{x\neq 0\}$.

$X$ is Hausdorff, but $Y$, the "line with doubled origin", is locally Hausdorff and therefore $T_1$, but not Hausdorff.

In fact, every topological space is the quotient of a Hausdorff space. This appears to have been proved by M. Shimrat in 1956, though there may be a very short argument.

I think that something stronger should be true: every topological space $Y$ is a quotient of a Stonean space, which is a Hausdorff extremally disconnected space. I believe that it is possible to show this by embedding the complete lattice of opens in $Y$ into a complete Boolean algebra, though if $Y$ is not sober this may complicate the proof or weaken the conclusion.

Andrew Dudzik
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  • @CameronBuie Ah, I seem to have forgotten that rather crucial observation. – Andrew Dudzik Dec 06 '15 at 17:39
  • I've been there! – Cameron Buie Dec 06 '15 at 17:45
  • thanks, can i ask what gave you the motivation for this example? – Matan L Dec 06 '15 at 17:50
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    @MatanL I first picked the standard $T_1$ example of the cofinite topology on an infinite set, but I could not quickly determine if it was the quotient of a Hausdorff space. So I tried picking a space with stronger properties, and the line with doubled origin is the standard example of a non-Hausdorff locally Hausdorff space—and it is quickly intuitive that it is a quotient of a Hausdorff space. I recently asked a question about locally Hausdorff spaces, so they've been on my mind. – Andrew Dudzik Dec 06 '15 at 17:56
  • again, thanks :) – Matan L Dec 06 '15 at 17:58
  • @MatanL Sure thing. :) I just added some musings; it seems to be the case that every topological space is a quotient of a Hausdorff space, though perhaps not trivially so. – Andrew Dudzik Dec 06 '15 at 18:16
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    There is a very short argument that every $T_1$ space is a quotient of an extremally disconnected Hausdorff space using convergence of ultrafilters (namely, $X$ is the quotient of the disjoint union of copies of spaces with the same underlying set as $X$, one witnessing each nonprincipal ultrafilter which converges in $X$). With a slight modification to handle principal ultrafilters, it can be adapted to non-$T_1$ spaces as well. – Eric Wofsey Dec 06 '15 at 18:20