Exercise · ZW-SEQUENTIAL-COMPACTNESS-METRIC-SPACES-006 · ★★☆ Intermediate

Let \(x_n=\cos\!\big(\tfrac{n\pi}{2}\big)\) in \([-1,1]\). Which values are limits of some subsequence?

Choose one option.

Show answer
  • B. \(\{-1,\,0,\,1\}\)

Explanation

The values of the sequence are \(0,-1,0,1\) repeated with period \(4\): each appears infinitely often, so each is the limit of a constant subsequence.

Study General topology in ZWMath →

Review the concept: Sequential compactness

A metric space \(X\) is sequentially compact if every sequence \((x_n)\) in \(X\) has a subsequence \((x_{n_k})\) converging to a point \(x\in X\).

\[\forall (x_n)\subseteq X\ \ \exists\, n_1

More practice problems: sequential compactness

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