I've seen some proofs by definition and just want to ask for proof verification on whether this is okay as well:

Given complete metric space $(X,d)$ and $A \subset X$, $A$ closed. Prove $A$ is complete.

We know any cauchy sequence in $X$ converges, i.e. $\{a_n\} \subset X$ and cauchy implies $a_n \to a \in X$.

Suppose for contradiction, $A$ is not complete, then $\exists \{a_n\} \subset A$ and cauchy such that $a_n$ does not converge in A. Since $A \subset X$, this sequence converges to $a \in X/A$.

Since $A$ closed, $X/A$ open, so $\exists r > 0$ s.t. $B_r(a) \subset X/A$. Then this contradicts that it is cauchy by picking any $\epsilon < r$.


2 Answers 2


This is basically fine, but you don't really need to go by contradiction. $A\subset X$ is closed iff it contains all of its limit points. Just let an arbitrary Cauchy sequence, $\{a_{n}\}\subset A$ be given, it must converge in $X$ by completeness, and then the closedness directly implies that this limit is in $A$, making $A$ a complete metric space in its own right.

  • $\begingroup$ I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of $\{a_n\}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :) $\endgroup$ Commented Nov 28, 2018 at 6:30
  • $\begingroup$ No worries! Keep up the good work! $\endgroup$
    – JWP_HTX
    Commented Nov 28, 2018 at 6:48

Sure, this proof works!

Just a few comments, though: firstly, take the sentence

We know any cauchy sequence in $X$ converges, . . .

In my opinion, it would be preferable to say that

We know any Cauchy sequence in $X$ converges in $X$, . . .

just for the sake of more clarity, even though you have explained what this means by adding an "i.e.".


Then this contradicts that it is cauchy by picking any $\epsilon < r$.

It will be helpful to explain how the contradiction arises. At first glance, I can see that there will be a contradiction to the fact that $\{ a_n \}$ converges to $a$, not that $\{ a_n \}$ is Cauchy. But, the two ideas are closely related enough that I am still confident that you're not necessarily wrong. So, adding more details here would be better, in my opinion.


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