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$(X,\mathcal T)$ is a topological space which has a dense Hausdorff subspace. Is $X$ Hausdorff?

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5 Answers 5

up vote 23 down vote accepted

For a generalisation, one can easily show that every Hausdorff space can be embedded as a dense subset of a non-Hausdorff space. Given a Hausdorff space $X$, define a new space $Y = X \cup \{ \alpha , \omega \}$ so that the open sets are just the open sets of $X$ together with the whole space $Y$. Then $X$ is clearly a dense subspace of $Y$ (the only neighbourhood of either $\alpha$ or $\omega$ is the entire space $Y$), and $Y$ is not Hausdorff (actually, it is not even T$_0$, since $\alpha$ and $\omega$ have exactly the same open neighbourhoods).

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thanks‌‌‌‌‌‌‌‌. –  user59671 Mar 4 '13 at 18:43
    
@CutieKrait: You're welcome! I thought a simple general construction would be a nice addition to the answers already given. –  Arthur Fischer Mar 4 '13 at 18:50
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No.

Consider $\Bbb R$ with the trivial topology $\{0\}$ is a dense subset, it is certainly Hausdorff, but space itself is certainly not.

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For a slightly less trivial example, consider $X = [0,1] \cup \{1'\}$, where $1'$ is an "extra copy of $1$" so that basic neighbourhoods of $1'$ are $\{1'\} \cup (1-\epsilon,1)$ for any $\epsilon > 0$. On $[0,1]$, the usual topology is taken. Then $[0,1]$ is a dense subspace which is Hausdorff, but $X$ is not Hausdorff because $1$ and $1'$ don't have disjoint neighbourhoods.

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Algebraic geometers will smile and think of the generic point of their favourite irreducible scheme...

[For non algebraic geometers: a scheme is the basic object studied in algebraic geometry.
It has an underlying topological space which is almost never Hausdorff.
Very often however a scheme $X$ has a so-called generic point $\eta\in X$, such that the singleton set $\{\eta\}$, obviously a Hausdorff subspace, is dense in $X$. The scheme $X$ is then called irreducible.
I am answering the question in order to show that the situation arises in real (mathematical) life, contrary to what some (excellent) other answers might suggest.]

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The trivial topology arises more often than one thinks! It's one of my favorite sources for counterexamples! :-) –  Asaf Karagila Mar 4 '13 at 23:39
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No, Sierpiński space is the closure of its non-closed point.

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