Can anyone give an example of a non-Hausdorff space that contains a Hausdorff subspace?


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A point is Hausdorff, so any non-Hausdorff space gives examples.

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    $\begingroup$ Might as well take the empty subspace of any non-Hausdorff space. $\endgroup$ – Ittay Weiss Mar 10 '17 at 20:37
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    $\begingroup$ Wow, really? Who doesn't allow the empty space? That seems totally unnatural. $\endgroup$ – Kevin Carlson Mar 10 '17 at 21:20
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    $\begingroup$ @KevinCarlson. I agree. Every text or paper I've read allows the empty set to be a top. space . It would be awkward if it weren't. It would be like allowing $1$ to be prime: Almost every mention of primes would have to be changed to "primes greater than $1$". $\endgroup$ – DanielWainfleet Mar 10 '17 at 21:50
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    $\begingroup$ @user254665 the situation is the opposite here though, allowing the empty space means that some mentions of "topological space" have to be changed to "non empty topological space", see for example the statement of Banach fixed point theorem, or the assertion "every non empty complete metric space without isolated points is uncountable" to mention the firsrt 2 that came to my mind. (That doesn't mean we shouldn't allow the empty space, just that I don't see your point) $\endgroup$ – Alessandro Codenotti Mar 11 '17 at 0:27
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    $\begingroup$ @AlessandroCodenotti: How about "the intersection of two subspaces is a subspace", or "every set can be given the discrete topology", or "the fiber over a point $b$ of a bundle $E \to B$" or "given a continuous map $X \to Y$, the inverse image of an open subspace is an open subspace". These basic notions all become very awkward when you exclude the empty space. $\endgroup$ – Hurkyl Mar 11 '17 at 3:14

It is enough to consider the line with two origins, and than consider one of this lines as a subspace.

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    $\begingroup$ For those who don't recognize it from the description, add an additional point $0'$ to the real numbers and take the sets $(-\epsilon, 0)\cup\{0'\}\cup(0,\epsilon)$ to be a basis of neighborhoods of $0'$. $\Bbb R$ is a subspace, but $0$ and $0'$ are not separable. $\endgroup$ – Paul Sinclair Mar 11 '17 at 6:40

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