Why is $[-1,1]$ compact when $a_n = (-1)^k$ does not converge in $A$

Question

I know this question sounds silly but I was reading the definition of compactness and couldn't quite wrap my head around this

Compactness :A subset $A$ of a metric space $M$ is compact if every sequence $(a_n)$ in $A$ has a subsequence $(a_{n_k})$ that converges to a limit in $A$

• Let $A = [-1,1]$, then a sequence in $A$ is $a_n = (-1)^k$ which oscillates between $-1$ and $1$

  Any sequence is a subsequence of itself

  $a_n = (-1)^k$ does not converge in $A$, therefore $A$ is not compact

Can someone point out my error please!

Answer 1

It's not silly question. The part you didn't understand is: "has a subsequence"= "has at least one" - that means it is enough to have just one subsequence that converges in $A$, for example $a_{2k}$ but all others don't have to converge, for example sequence $a_k$ as subsequence itself.

Maybe counterexample will help you to understand more:

You can use the same theorem to prove that $B=\langle0,1]$ is not compact since $b_n=(\frac{1}{n})$ is sequence in $B$ which is convergent in $\mathbb{R}$ with limit $0$. And you know that every subsequence of convergent sequence has the same limit, $0$ in this case, and since $0\notin B$ there is no subsequence of $b_n$ that converges in $B$ so $B$ is not compact.

Answer 2

Every sequence needs to have a converging subsequence but not every subsequence needs to converge. In your case obviously $a_{2k}$ converges which is sufficient to see that the sequence is no counter example.

Answer 3

You are confusing the quantifiers. It says that for every sequence ($\forall$), there exists ($\exists)$ a subsequence which converges.

That means in your space, pick any arbitrary sequence. Then if the space is compact, there should exist one (or more) convergent subsequences. Contrast with saying for every sequence and every subsequence of the sequence converges.

In your example, you've picked $((-1)^k)$ so what is the converging subsequence? In fact, you can find more than one: by taking $k$ to be just even or just odd, then the subsequences will be constant at $1$ and $-1$ respectively.