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Let $(X, d)$ be a metric space and let $\{x_n\}$ be a convergent sequence in $X$ with limit $x$.

I am having difficulty proving that $A = \{x_n\} \cup \{x\}$ is a (sequentially) compact set. That is, every sequence $\{y_n\} \subseteq A$ contains a convergent subsequence.

Navies
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  • Why not just prove directly that $A$ is compact? That’s very easy, and in a metric space compactness is equivalent to sequential compactness. – Brian M. Scott Oct 19 '16 at 20:43
  • By compact, do you mean that every open cover of $A$ admits a finite subcover?. We have no covered that definition in class yet and have only seen the definition of sequentially compactness in a set. – Navies Oct 19 '16 at 20:47
  • That’s seems a rather odd approach to the subject. Oh, well. Give me a few minutes, and I’ll write up an answer to point you in the right direction. – Brian M. Scott Oct 19 '16 at 20:50
  • Since @BrianM.Scott is writing something up then I will not write anything more. However, you should try to prove the statement via contradiction. – Jacky Chong Oct 19 '16 at 20:51
  • @Navies what is your textbook in this course? – Hamit Oct 19 '16 at 21:14
  • We are most following the course notes prepared by the professor. – Navies Oct 19 '16 at 21:20

1 Answers1

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HINT: Let $\sigma=\langle y_k:k\in\Bbb N\rangle$ be a sequence in $A$.

  • If there is some $p\in A$ such that $\{k\in\Bbb N:y_k=p\}$ is infinite, show that $\sigma$ has a constant subsequence converging to $p$.

Otherwise, show that there is a subsequence $\langle y_{n_k}:k\in\Bbb N\rangle$ of distinct points of $A$. At most one of these points is $x$; removing it still leaves us with a subsequence of $\sigma$ consisting of distinct points, so we might as well assume that $y_{n_k}\ne x$ for each $k\in\Bbb N$. This means that for each $k\in\Bbb N$ there is a unique $f(k)\in\Bbb N$ such that $y_{n_k}=x_{f(k)}$, so that

$$\langle y_{n_k}:k\in\Bbb N\rangle=\langle x_{f(k)}:k\in\Bbb N\rangle\;.\tag{1}$$

If the sequence $\langle f(k):k\in\Bbb N\rangle$ were increasing, the sequence in $(1)$ would be a subsequence of $\langle x_n:n\in\Bbb N\rangle$ and would therefore converge to $x$. Unfortunately, $\langle f(k):k\in\Bbb N\rangle$ need not be an increasing sequence.

  • Show that $\langle f(k):k\in\Bbb N\rangle$ has an increasing subsequence $\langle f(k_i):i\in\Bbb N\rangle$. Conclude that $\langle y_{n_{k_i}}:i\in\Bbb N\rangle=\langle x_{f(k_i)}:i\in\Bbb N\rangle$ is a subsequence of $\langle x_n:n\in\Bbb N\rangle$ and therefore converges to $x$.
Brian M. Scott
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