# Is there a metric space such that every Cauchy sequence in it does not converge?

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?

• Yes: you are correct. All constant sequences must converge, and are obviously Cauchy. Dec 6, 2019 at 22:48
• 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. Dec 6, 2019 at 22:49
• Let $(X,d)$ be a metric space. $X$ is discrete iff Cauchy sequences in $X$ converge only when they are eventually constant. Dec 6, 2019 at 22:56

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.

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