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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 $x\neq y$ ?

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

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  • $\begingroup$ $N$ contains a neighbourhood of $(x,x)$. $\endgroup$ – Lord Shark the Unknown Mar 3 at 7:55
  • $\begingroup$ @LordSharktheUnknown, Hence we can say that $x\neq y$. Is it true? $\endgroup$ – user479859 Mar 3 at 7:58
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Mar 3 at 8:46
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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 $x \neq y$.

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