Show that a vector subspace is a closed subset of \(\mathcal{H}\) with respect to the norm topology iff the limit of every sequence of vectors in \(V\) belongs to \(V\).

Start by defining the crucial terms. A vector subspace \(V\) of a vector space \(\mathcal{H}\) is a set of vectors that is closed under vector addition and scalar multiplication. This means that for any two vectors \(v_1, v_2 \in V\), the sum \(v_1 + v_2\) is also in \(V\), and for any scalar \(\alpha\) and vector \(v \in V\), the product \(\alpha v\) is also in \(V\). In the context of topological spaces, a set is considered closed if its complement is open. In the norm topology of a Hilbert space, a set is open if for each point in the set, there exists an open ball contained in the set. The limit of a sequence in a set \(V\) belongs to \(V\) iff for any \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that for all \(n > N\), the distance between the \(n\)th term of the sequence and the limit is less than \(\epsilon\).

Suppose \(V\) is a vector subspace of \(\mathcal{H}\) such that the limit of every sequence in \(V\) belongs to \(V\). Now, to show that \(V\) is closed, it needs to be shown that the compliment of \(V\), say \(\mathcal{H} - V\), is open in \(\mathcal{H}\). Take an arbitrary point \(h \in \mathcal{H} - V\). If \(\mathcal{H} - V\) is not open, then there would exist a sequence of vectors in \(V\) converging to \(h\). But this would contradict the hypothesis that V is closed under limits, so \(\mathcal{H} - V\) is indeed open and thus \(V\) is closed.

Assume \(V\) is a closed vector subspace of \(\mathcal{H}\). Suppose by contradiction that there exists a sequence \((v_n)\) in \(V\) whose limit \(v\) does not belong to \(V\). In other words, \(v \in \mathcal{H} - V\) (the complement of \(V\) in \(\mathcal{H}\)). But since \(V\) is closed, \(\mathcal{H} - V\) is open. As \(v \in \mathcal{H} - V\), there exists an open ball centered at \(v\) which does not intersect \(V\). This contradicts that \(v\) is the limit of a sequence in \(V\). Thus every limit of sequence in \(V\) must belong to \(V\).