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Suppose we have a metric space $(X,d)$, can $x_n \to x' $ as $n\to \infty$ such that $x'\notin X$ for any Cauchy sequence if the interior of $X$ is nonempty?

My logic: if $X$ has nonempty interior, then we can define a constant sequence $x_n = x$ with $x\in X$. And this constant sequence is Cauchy obviously. Hence, there is no such metric space.

Is my logic correct?

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    $\begingroup$ Yes: you are correct. All constant sequences must converge, and are obviously Cauchy. $\endgroup$
    – Crostul
    Dec 6, 2019 at 22:48
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    $\begingroup$ It doesn't make sense to say $x_n \rightarrow x'$ if $x' \notin X$ unless we are considering $X$ as a subset of some larger space. $\endgroup$
    – Jair Taylor
    Dec 6, 2019 at 22:49
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    $\begingroup$ Let $(X,d)$ be a metric space. $X$ is discrete iff Cauchy sequences in $X$ converge only when they are eventually constant. $\endgroup$
    – Hagen von Eitzen
    Dec 6, 2019 at 22:56

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You are right. No such non-empty metric space exists. Let $(X,d)$ be an arbitrary non-empty metric space.

Fix $x \in X$ and define a sequence $(x_n)_{n=1}^\infty$ by $x_n:= x, n \geq 1$. Then clearly $\lim_{n \to \infty} x_n = x$ and the sequence $(x_n)_{n=1}^\infty$ is a convergent/Cauchy sequence.

But what happens when $X = \emptyset$?

Then vacuously, every Cauchy sequence in $X$ does NOT converge (but also converges!). So the only such metric space is the empty metric space.

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  • $\begingroup$ Wow the last statement is so weird. I immediately thought it was a contradiction, but I guess its not. $\endgroup$
    – Ovi
    Dec 6, 2019 at 23:06
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    $\begingroup$ Haha yes, you can say that every Cauchy sequence in the empty set both converges and not converges. $\endgroup$
    – J. De Ro
    Dec 6, 2019 at 23:07
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    $\begingroup$ +1 for the beautiful last paragraph! $\endgroup$
    – Severin Schraven
    Dec 6, 2019 at 23:12

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