Exercise · ZW-SEQUENTIAL-COMPACTNESS-METRIC-SPACES-004 · ★☆☆ Basic
In \(X=(1,2]\) with the usual distance, does the sequence \(x_n=1+\tfrac1n\) have a subsequence converging in \(X\)?
Choose one option.
Show answer
- D. No: every subsequence tends to \(1\), which is not in \(X\)
← Back to the question
Explanation
The sequence is bounded and converges to \(1\), so all its subsequences tend to \(1\notin(1,2]\). That is why \((1,2]\) is not sequentially compact.
A metric space \(X\) is sequentially compact if every sequence \((x_n)\) in \(X\) has a subsequence \((x_{n_k})\) converging to a point \(x\in X\).