Apologies for my title. I couldn't get all the info in without running out of room. Feel free to edit. I'm just looking for verification of my proof. This problem is from some study materials I am working through, and my solution feels a little too easy. I will state the problem verbatim below, and also provide an answer for critiquing. If it has flaws, I will delete my answer. Others can feel free to answer either way.

Problem: Let $\{x_n\}$ be a sequence in a complete metric space $(X,d)$ and for every $\varepsilon>0$ there exists a convergent sequence $\{y_n\}$ such that $\sup_nd(x_n,y_n)<\varepsilon$. Prove that then $\{x_n\}$ converges.

Here is the answer I came up with:

Let $\varepsilon>0$ be given. Chose a sequence $\{y_n\}$ such that $y_n\to y$ for some $y\in X$ and $\sup_{n\in\mathbb{N}}d(x_n,y_n)<\frac{\varepsilon}{2}$. Such a sequence exists by hypothesis. Since $y_n\to y$, there exists $N\in\mathbb{N}$ such that $\forall n\geq N$, $\;d(y_n,y)<\frac{\varepsilon}{2}$. Then, $\forall n\geq N$, we have by the triangle inequality that:

$$d(x_n,y)\leq d(x_n,y_n)+d(y_n,y)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$

and since $X$ is complete, we have that $x_n\to y$, and $\{x_n\}$ is convergent in $(X,d)$.

  • $\begingroup$ For the proof-verification, you probably should have put your "proof to be checked" in the question, not as answer. $\endgroup$ – Clement C. Jan 18 '17 at 18:34
  • $\begingroup$ @ClementC. Oh, good point. Noted. $\endgroup$ – The Count Jan 18 '17 at 18:35
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    $\begingroup$ Can the downvoter explain? $\endgroup$ – The Count Jan 26 '17 at 14:22
  • $\begingroup$ I'm afraid most of them don't care enough to (in this case, I really don't see a reason to downvote). $\endgroup$ – Clement C. Jan 26 '17 at 14:31
  • $\begingroup$ @ClementC. I know who it was. They downvote me once every few days because I have called them out for never accepting answers, sloppy editing, and a general poor attitude. I'm just getting the vibe out there. $\endgroup$ – The Count Jan 26 '17 at 14:34

The "bug."

The issue with your proof is that $y$ depends on $\varepsilon$ itself, so the end statement you show has the quantifiers wrong: you show "$\forall \varepsilon > 0,\ \exists y,\ [\cdots]$" instead of "$\exists y,\ \forall \varepsilon > 0,\ [\cdots]$."

One thing that should be alarming is that nowhere have you actually used the completeness of $X$. You wrote "since $X$ is complete," but you are not using it to argue anything. But one rarely throw in unnecessary assumptions for the sake of it.

A proof.

Let $\varepsilon > 0$ be an arbitrary positive number. Choose a convergent sequence $(y^{(\varepsilon)}_n)_{n\geq 0}$ such that $$\sup_{n\geq 0} d(x_n,y^{(\varepsilon)}_n) \leq \frac{\varepsilon}{3} \tag{1} $$

Since $(y^{(\varepsilon)}_n)_{n\geq 0}$ is convergent, it is Cauchy: let $n_\varepsilon\geq 0$ be such that for all $n\geq n_\varepsilon$ and $m\geq 0$, $$d(y^{(\varepsilon)}_{n+m},y^{(\varepsilon)}_n) \leq \frac{\varepsilon}{3}\tag{2}$$

Now, for any $n\geq n_\varepsilon$ and $m\geq 0$, using (1), (2), and the triangle inequality $$ d(x_{n+m},x_n) \leq d(x_{n+m},y^{(\varepsilon)}_{n+m}) + d(y^{(\varepsilon)}_{n+m},y^{(\varepsilon)}_n) + d(y^{(\varepsilon)}_n,x_n) \leq \frac{\varepsilon}{3} +\frac{\varepsilon}{3} +\frac{\varepsilon}{3} = \varepsilon $$ showing that $(x_n)_{n\geq 0}$ is Cauchy. By completeness of $X$, it it thus convergent.

  • $\begingroup$ Thanks very much, Clement. I'll go through this in detail in a little while and make sure it makes sense to me. $\endgroup$ – The Count Jan 18 '17 at 18:49
  • $\begingroup$ Having read it, so my mistake was trying to relate the sequence $\{x_n\}$ to the point that $\{y_n\}$ converges to? It looks you did what I did but just used the Cauchy-ness of the $y_n$ sequence instead. Am I understanding correctly? $\endgroup$ – The Count Jan 18 '17 at 19:05
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    $\begingroup$ Yes, exactly. ${}$ $\endgroup$ – Clement C. Jan 18 '17 at 19:06

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