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Considering the seguent proposition:

Suppose that $(X,T)$ is a topological space. Show that if $X$ is Hausdorff, the diagonal $Δ=\{(x,x)∣x∈X\}$ is closed in $X×X$.

Now I did an example (probably wrong) where I show that $\Delta$ is closed and $X×X$ it's not an Hausdorff space.

Let $T=\{\emptyset, \mathbb{R^2}\} \cup \{\mathbb{R^2}\setminus \Delta\} $. So $\Delta$ is a closed set. If I take two point $x,y\notin \Delta$ the unique open set that contain $x$ and $y$ is $\mathbb{R^2}\setminus \Delta $. So $X×X$ can't be a Hausdorff space then $X$ can't be a Hausdorff space.

Where is the mistake? Thank you for helps

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  • $\begingroup$ it's not a duplicate! I tryed to show a contradction which proved to be false @NajibIdrissi $\endgroup$ – Skills Feb 21 '15 at 12:51
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You don't define a topology on $\mathbb{R}$. You define one directly on the square.

The theorem states that if we have a topology on $X$ and consider $\Delta$ in $X \times X$ in the product topology. If $X$ is Hausdorff then $\Delta$ is closed in $X \times X$ (in the product topology!).

The reverse also holds: if $\Delta$ is closed, $X$ is Hausdorff.

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