Chapter 10: Problem 30

Show that if \(x_{n}\) is a sequence in anormed vector space \(V\) such that every subsequence has a subsequence comergent to \(x\), then \(x_{n} \rightarrow x\).

Short Answer

Expert verified

If any subsequence of \(x_n\) has a subsequence convergent to \(x\), then \(x_n\) converges to \(x\). This is proven by constructing a subsequence that for each positive integer \(m\), \(\|x_{n_k} - x\| < 1/m\) for \(k > m\). The index in this sequence can be used to demonstrate \(x_n\) satisfies the conditions of convergence.

Step by step solution

State The Property to Be Shown

To show that \(x_n \rightarrow x\), it has to be proven that for any given \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that \(\| x_n - x \| < \epsilon\) for all \(n > N\). This is the definition of convergence in normed vector spaces.

Construct Subsequence

For each positive integer \(m\), choose a subsequence \(x_{n_k}\) such that \(\|x_{n_k} - x\| < 1/m\) for all \(k>m\). It is possible to do so since every subsequence of \(x_n\) must have a convergent subsequence, which by definition means an \(N\) exists such that \(\|x_{n_k} - x\| < 1/m\) for all \(k>N\). Choose the first element of the subsequence where this inequality holds.

Use Subsequence to Prove Convergence of Original Sequence

The constructed subsequence the properties that for a given positive integer \(m\), the index \(n_k\) associated with the constructed subsequence is such that for all \(n > n_k\), \(\|x_n - x\| < 1/m\). This shows that for any given \(\epsilon = 1/m\), an integer \(N = n_k\) can be found, such that for all \(n > N\), \(\|x_n - x\| < \epsilon\). Therefore, we have shown that \(x_n\) converges to \(x\).