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Let $(x_{n})_{n=m}^{\infty}$ be a sequence in $(X,d)$ which converges to some limit $x_{0}$. Then every subsequence $(x_{f(n)})_{n=m}^{\infty}$ of that sequence also converges to $x_{0}$.

My solution

Let $\varepsilon > 0$. Then there exists a natural number $N\geq m$ such that \begin{align*} n\geq N \Rightarrow d(x_{n},x_{0}) < \varepsilon \end{align*}

Since $f:\textbf{N}\to\textbf{N}$ is stricly increasing, we conclude that $f(n) \geq n$.

Consequently, for the same $N\geq m$, the following relation holds \begin{align*} f(n) \geq n\geq N \Rightarrow d(x_{f(n)},x_{0}) < \varepsilon \end{align*}

whence we conclude that $x_{f(n)}$ converges to $x_{0}$ as well.

I am mainly concerned with the wording of the proof. Can someone point out any flaw?

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Your proof (and the wording as well) is fine. You proved that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $n \geq N \implies x_{f(n)} \in B(x_0, \varepsilon)$, which shows exactly the convergence you desired.

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While "$f$ is strictly increasing implies $f(n) \ge n$" is true, if I were grading this as a homework problem I'd like to see this proved, and not just concluded. Can you prove it? Give it some thought, and if you're stuck I've given a hint below as to how I'd try. (It's protected as a spoiler. Just mouse over it to reveal.)

Induction

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  • $\begingroup$ Good point. Would this suffice as a proof? Suppose $f(n) > n$. Then since $f$ is increasing, $f(n+1) > f(n) > n$. So, $f(n+1) - n > 0$. But since $f(n+1)$ and $n$ are both natural numbers, this is the same as $f(n+1) - n \geq 1$. Rearranging this inequality proves the result. $\endgroup$ – Nicholas Roberts May 24 at 3:49
  • $\begingroup$ What are you worried about, exactly? This seems like weird advice. I thought that was already an oddly overwrought and unnecessary part of the proof, since the standard definition of a subsequence is, for example, "A subsequence of $\{a\}$ is a sequence $\{b\}$ defined by $b_k=a_{n_k}$, where $n_1<n_2<... $ is an increasing sequence of indices (D'Angelo and West 2000)." (from mathworld.wolfram.com/Subsequence.html ) . Since $n_k$ is an infinite subset of $\mathbb{N}$, considering the tail of the subsequence such that $n_k$ is greater than some finite $N$ is fine. $\endgroup$ – user762914 May 24 at 3:52
  • $\begingroup$ @NicholasRoberts: Yep, that's how I'd do it. Small correction in that you should write "$f(n) \ge n$", not "$f(n) \gt n$". But the argument still works with that change. However you should also read Renard's comment, which I think is valid. $\endgroup$ – JonathanZ supports MonicaC May 24 at 14:20
  • $\begingroup$ @Renard: The OP asked for a critique of their solution, so they are clearly just learning how to write up proofs. Being able to recognize the difference between asserting something because it makes sense and actually proving it is a hard thing to learn, and their use of "$f(n) \ge n$" was one where they had not proved it. That's what I was worried about. However .....(cont'd) $\endgroup$ – JonathanZ supports MonicaC May 24 at 14:27
  • $\begingroup$ However, I agree with you that that that was the most "off" part of the proof. Your use of "subsequence is guaranteed to get past and stay past $N$" seems much more sensible, and much better reflects the spirit of a subsequence.(cont'd) $\endgroup$ – JonathanZ supports MonicaC May 24 at 14:35

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