2
$\begingroup$

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 $\epsilon < r$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 :) $\endgroup$ – OneRaynyDay Nov 28 '18 at 6:30
  • $\begingroup$ No worries! Keep up the good work! $\endgroup$ – JWP_HTX Nov 28 '18 at 6:48
2
$\begingroup$

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.".

Secondly,

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.