Let \(\ell_{0}\) be the subset of \(\ell^{2}\) consisting of sequences with only finutely many terms different from zero. Show that \(\ell_{0}\) is a vector subspace of \(\ell^{2}\), but that it is not closed. What is its closure \(\overline{\ell_{0}} ?\)

To prove that a set \(\ell_{0}\) is a subspace of \(\ell^{2}\), it needs to satisfy three properties: i) \(\ell_{0}\) is non-empty. ii) If \(u, v\) are in \(\ell_{0}\), then \(u + v\) is in \(\ell_{0}\). iii) If \(u\) is in \(\ell_{0}\) and \(c\) is any scalar, then \(c \cdot u\) is in \(\ell_{0}\). It's clear that \(\ell_{0}\) is non-empty since it consists of sequences with only finitely many terms different from zero. If \(u, v\) are in \(\ell_{0}\), then \(u + v\) are also in \(\ell_{0}\) because addition of two sequences with finitely many non-zero terms also results in a sequence with finitely many non-zero terms. Also, if \(u\) is in \(\ell_{0}\) and \(c\) is a scalar, then \(c \cdot u\) is in \(\ell_{0}\) since multiplication of a sequence with finitely many non-zero terms by a scalar also results in a sequence with finitely many non-zero terms. Hence, \(\ell_{0}\) satisfies all the properties of a vector subspace.

A set is called closed if it contains all its limit points. Consider the sequence of sequences \((x^{n})\) where \(x^{n}_{i}=0\) for \(i \neq n\) and \(x^{n}_{n}=1\). Every sequence \((x^{n})\) is clearly in \(\ell_{0}\), but the limit of \((x^{n})\) as \(n\) approaches infinity is the sequence \(y\) where \(y_{i}=0\) for all \(i\), which is not in \(\ell_{0}\). This shows that \(\ell_{0}\) does not contain all its limit points and therefore is not closed.

The closure of \(\ell_{0}\), denoted by \(\overline{\ell_{0}}\), is the smallest closed set that contains \(\ell_{0}\). In this case, \(\overline{\ell_{0}}\) is the entire space \(\ell^{2}\) since every sequence in \(\ell^{2}\) can be approximated by a sequence in \(\ell_{0}\).