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It should be a triviality, I believe.

The topology induced by a uniform structure $\mathcal{U}$ with $\cap \mathcal{U} =\Delta$, where $\Delta$ is the diagonal, is Hausdorff.

Now I think that if I define the product topology $X \times X$ to be such that the closed sets of it are the sets that are in the $\mathcal{U}$, then I will get that $\Delta$ is closed as an arbitrary intersection of closed sets, and thus the space $X$ is $T_2$.

Is this right? (I am just not sure that if I induce the sets in $\mathcal{U}$ as open sets in $X \times X$ that the above will occur).

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If you define a topology on $X\times X$, there is no reason why this topology will have anything to do with a topology on $X$. – Mariano Suárez-Alvarez Nov 16 '11 at 16:03
up vote 2 down vote accepted

You don’t need to look at any topology on $X\times X$; just show Hausdorffness of $X$ directly. Let $x$ and $y$ be distinct points of $X$. Then $\langle x,y\rangle\notin\Delta$, so there is some $U\in\mathcal{U}$ such that $\langle x,y\rangle\notin U$. Now use the uniformity axioms to show that there is a symmetric $V\in\mathcal{U}$ such that $V\circ V\subseteq U$, and then show that $V[x]$ and $V[y]$ must be disjoint nbhds of $x$ and $y$, respectively.

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I am not sure, but the symmetric set $V$ I am looking is actually $V=U\cap U^{-1}$, right? I can see that it equals $V^{-1}$, cause $$(x,y)\in V \leftrightarrow (x,y) \in U & (y,x) \in U$$ and $$(x,y) \in V^{-1} \leftrightarrow (y,x) \in U & (x,y) \in U$$. I see now that it's really as easy as the lecturer said. :-) – MathematicalPhysicist Nov 16 '11 at 17:37
@MathematicalPhysicist: Whether you can use $U\cap U^{-1}$ directly depends on exactly how the axioms of a uniformity were given. It’ll work with the definition here, for instance. Definition 35.2 in Willard’s General Topology, on the other hand, requires a couple of extra steps: it guarantees a $V_1\in\mathcal{U}$ with $V_1\circ V_1\subseteq U$ and then a $V_2\in\mathcal{U}$ such that $V_2^{-1}\subseteq V_1$; then $(V_1\cap V_2)\cap(V_1\cap V_2)^{-1}$ works. – Brian M. Scott Nov 16 '11 at 17:59
In the definition we were given there wasn't mention of condition d) in def 35.2 of Willards' text. This added condition makes the definition inequivalent, right? – MathematicalPhysicist Nov 18 '11 at 8:39
@MathematicalPhysicist: I’d have to see your definition to be sure. I suspect that yours actually implies that condition. – Brian M. Scott Nov 18 '11 at 10:35

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