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Can a topological space be Hausdorff and separable, but neither Lindelof nor first countable?

Can a topological space be Hausdorff and Lindelof, but neither separable nor first countable?

Can a topological space be Hausdorff, but not separable, Lindelof, or first countable?

Are all second countable spaces Hausdorff?

Sorry for posting several questions in one post, but I am just trying to figure out how the Hausdorff property relates to Lindelof, Separability, and First Countability in topological spaces.

Martin Sleziak
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Almangiekey
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  • The last question is particularly easy (at least in the sense that I already know the answer): let $X = \mathbb{R} \times {0,1}$ (where ${0,1}$ has the discrete topology) modulo the equivalence relation generated by $(x,0) \sim (x,1)$ for all $x \in \mathbb{R} \setminus {0}$. Then $X$ is second-countable but not Hausdorff ($X$ is known as the "line with doubled origin"). – diracdeltafunk Sep 18 '19 at 00:19
  • Or even easier: the trivial (indiscrete) topology on any set. – Nate Eldredge Sep 18 '19 at 01:39
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    @NateEldredge . As a professional nitpicker, I must point out that the trivial topology on a set with less than 2 members is the only topology on it, and it IS Hausdorff. – DanielWainfleet Sep 19 '19 at 00:21
  • Separable & Lindelof & 2nf-countable are "global" properties but 1st-countable is "local". If you take the topological sum of a large family of spaces you can get a space with arbitrarily large "global" values without disturbing the "local" values of the members of the family. You can find much about this in various places in General Topology by R. Engelking (which has a thorough index) under the topic Cardinal Functions. – DanielWainfleet Sep 19 '19 at 13:43

2 Answers2

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For questions like this, you probably want to get a copy of the book Counterexamples in Topology, by Steen and Seebach. The "general reference chart" in the back of that book allows you to search the counterexamples by combinations of properties they do and don't satisfy.

Fortunately, this chart (and more) now exists in an online searchable form: $\pi$-Base.

Can a topological space be Hausdorff and separable, but neither Lindelof nor first countable?

Yes: The strong ultrafilter topology.

Can a topological space be Hausdorff and Lindelof, but neither separable nor first countable?

Yes: Many examples, including the ordinal space $[0,\omega_1]$ and the countable complement extension topology (the topology generated by the standard topology on $\mathbb{R}$ together with the countable complement topology on $\mathbb{R}$).

Can a topological space be Hausdorff, but not separable, Lindelof, or first countable?

Yes: Many examples, including the product topology on a product of uncountably many copies of an infinite discrete set.

Are all second countable spaces Hausdorff?

No. Many examples, including the cofinite topology on $\mathbb{N}$.

Steven Clontz
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Alex Kruckman
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  • +1.... For the 3rd Q we can also, from the 1st or 2nd Q take a Hausdorff space X that's not 1st-countable and then take the topological sum of uncountably many copies of X. – DanielWainfleet Sep 19 '19 at 00:29
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Looking in large products (easy way to get non-first countable spaces with varying properties):

  • $\mathbb{R}^\mathbb{R}$ in the product topology is separable, Hausdorff (even Tychonoff), but not Lindelöf, nor first countable.

  • $[0,1]^I$ (product topology) is not separable if $|I| > |\mathbb{R}|$, Hausdorff, compact (so certainly Lindelöf) and not first countable for uncountable $I$.

  • $\mathbb{R}^I$ with $|I| > |\mathbb{R}|$ is Tychonoff, not separable, not normal (so certainly not Lindelöf as Lindelöf plus regular implies normal) and not first countable.

The final question on non-Hausdorff spaces is different (most "natural" examples as above are Tychonoff, so Hausdorff), and there even the indiscrete topology suffices as an example, which is trivially second countable.

Henno Brandsma
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    +1... For Q # 3, the box-product topology on $\Bbb R^{\aleph_0}$ is another Tychonoff example. – DanielWainfleet Sep 19 '19 at 00:38
  • @DanielWainfleet Yes, nice one too. (We cannot eliminate Lindelöf via the normality route (at least, that's open in ZFC) but ${0,1}^{\aleph_0}$ is uncountable closed and discrete so that does it.). And then $|X|=\mathfrak{c}$ whch is a bonus). – Henno Brandsma Sep 19 '19 at 07:03
  • It seems oddly irritating to me that the normality of the box product topology on $\Bbb R^{\aleph_0}$ is still unknown. It just doesn't feel like a hard Q. Maybe someone will find a Forcing argument concerning it. – DanielWainfleet Sep 20 '19 at 00:47