# The direct sum of two spaces in Hilbert space

We know that if $H$ is a Hilbert space, and $H_1$ is a closed subspace, then $H$ can be decomposed as $H=H_1\oplus H_1^{\perp}$. I want to know the equality here is just in algebric sense, or also in topological sense? If in topological sense, is the direct sum of these two spaces equipped with product topology?

• The equality is algebraic. It means that every $x\in H$ can be written as $x=u+v$ where $u\in H_1$, $v\in H_1^{\perp}$. And the definition of $H_1^{\perp}$ is that it consist of all $x \in H$ that are orthogonal to everything in $H_1$. – DisintegratingByParts Mar 25 '17 at 0:55
• @TrialAndError: I would appreciate a comment on the answer I just left, since we both seem to have a different viewpoint on this matter – el_tenedor Mar 25 '17 at 21:33

From linear algebra we know, that every subspace $U$ of a vector space $V$ can be complemented such that $V = U \oplus U'$.
As you suspected, the topology is the product topology, since a sequence $x_n + y_n$ in $H_1 \oplus H_1^\perp$ converges iff $x_n$ and $y_n$ converges. Why? Because one fundamental property in Hilbert spaces (which guarantees complementability) is the fact that for each closed subspace $H_1$ there exists a continuous orthogonal projection $P$ onto it, i.e. the convergence of $x_n + y_n$ implies the convergence of $P(x_n + y_n) = x_n$ and of $(1 - P)(x_n + y_n) = y_n$.