Exercise · ZW-CONVERGENCE-IN-A-METRIC-SPACE-007 · ★★☆ Intermediate

Let \(X=(0,1)\) with the usual distance and \(x_n=\tfrac1{n+1}\). Does it converge in \(X\)?

Choose one option.

Show answer
  • C. No: its only candidate limit, \(0\), is not in \(X\)

Explanation

The sequence approaches \(0\), but \(0\notin X\). Convergence requires the limit to belong to the space.

Study General topology in ZWMath →

Review the concept: Convergence in a metric space

A sequence \((x_n)\) in a metric space \((X,d)\) converges to \(x\in X\) if, for every \(\varepsilon>0\), there is an index \(N\) from which \(x_n\) lies within distance \(\varepsilon\) of \(x\).

\[\forall\varepsilon>0\ \ \exists N:\ n\ge N\Rightarrow d(x_n,x)<\varepsilon\]

More practice problems: convergence in a metric space

← ZW-CONVERGENCE-IN-A-METRIC-SPACE-006 ZW-CONVERGENCE-IN-A-METRIC-SPACE-008 →