# Let $N$ be neighborhood of diagonal of $X$ $\Delta_X$. Is it true $int(N)\neq \emptyset$? [closed]

Let $$X$$ be a topological space and $$N$$ be a neighborhood of the diagonal $$X$$. Let $$\{x_n\}$$ and $$\{y_n\}$$ be in $$X$$ with $$(x_n, y_n)\notin N$$. If $$x_n\to x$$ and $$y_n\to y$$, is it true that $$x\neq y$$ ?

## closed as off-topic by GNUSupporter 8964民主女神 地下教會, Cesareo, José Carlos Santos, Vinyl_cape_jawa, Alex ProvostMar 3 at 16:34

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• $N$ contains a neighbourhood of $(x,x)$. – Lord Shark the Unknown Mar 3 at 7:55
• @LordSharktheUnknown, Hence we can say that $x\neq y$. Is it true? – user479859 Mar 3 at 7:58
• How is the title related to the actual question? If $N$ is a neighbourhood of the diagonal $\Delta_X$, it has non-empty interior in $X^2$ by definition. – Henno Brandsma Mar 3 at 8:46

We know that $$(x_n, y_n) \to (x,y)$$ in $$X\times X$$. Let $$O$$ be open in $$X \times X$$ with $$\Delta_X \subseteq O \subseteq N$$. By assumption, all $$(x_n, y_n) \in (X\times Y)\setminus O$$ (which is closed) and so $$(x,y) \in (X \times Y) \setminus O$$ and in particular $$(x,y) \notin \Delta_X$$ and $$x \neq y$$.