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Give an example of a topological space that is separable but not a Hausdorff space.

I have not been able to discover an example, I thought of the Arens-Fort space because this is separable, since this space is countable and the same set is a countable dense about yourself, but do not know how to show that is second-countable. This space serves as an example, there are some other more. Thanks

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    $\begingroup$ So by separable do you mean "has a countable dense subset" or "second countable?" Wouldn't any finite space that isn't Hausdorff be an example in either case? $\endgroup$ Jun 14, 2013 at 1:19
  • $\begingroup$ In this answer I describe the Arens-Fort Space, as well as demonstrate that it is normal (T$_4$), and thus Hausdorff. $\endgroup$
    – user642796
    Jun 14, 2013 at 3:06

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Just take a finite space that isn't Hausdorff, such as the set $X=\{s,\eta\}$ with the topology where the open sets are $X$, $\emptyset$, and $\{\eta\}$.

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  • $\begingroup$ aka the Sierpinski space $\endgroup$ Jun 14, 2013 at 2:32
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$\pi$-Base, which draws its data from Steen and Seebach's Counterexamples in Topology, lists the following separable, non-Hausdorff spaces. You can learn more about these spaces by viewing the search result.

Compact Complement Topology

Countable Excluded Point Topology

Countable Particular Point Topology

Deleted Integer Topology

Divisor Topology

Double Pointed Reals

Finite Complement Topology on a Countable Space

Finite Complement Topology on an Uncountable Space

Finite Excluded Point Topology

Finite Particular Point Topology

Hjalmar Ekdal Topology

Indiscrete Topology

Interlocking Interval Topology

Maximal Compact Topology

Nested Interval Topology

Odd-Even Topology

One Point Compactification fo the Rationals

Overlapping Interval Topology

Prime Ideal Topology

Right Order Topology on the Reals

Sierpinski Space

Telophase Topology

The Integer Broom

Uncountable Particular Point Topology

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  • $\begingroup$ Fantastic answer. Just let me point one example more. If I'm not wrong, the Arens' space presented here should be separable (take $D=\mathbb Z^+\times\mathbb Z^+$) but it shouldn't be Hausdorff (for arbitrary $n\in\mathbb Z$, $n$ and $\infty$ has no disjoint neighbouhoods). $\endgroup$
    – Dog_69
    Oct 13, 2018 at 13:59
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There are many examples. The simplest examples are as following:

1, Indiscrete Topology

2, Finite Complement Topology on a countably infinite set

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Since any set X is countable if there exists a bijective map from X into subset of N If m is fixed +ve integer such that (12345..........m) is that subset of N then m is the cardinality of X,nd X is countable. So here i forced to say that the finite complement topology on an countable (infinite) set is not Hausdorff . If X is finite then (X,Tf) is discrete topology so becomes Hausdorff.

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