Does the direct sum decomposition of a Hilbert space comprise of closed subsets?

Question

Let $X$ be a Hilbert space and suppose that

$$ X = Y \oplus Z,$$

where $Y, Z \subset X$. Is it necessarily the case that $Y$ and $Z$ must be closed subspaces of $X$?

Answer 1

No. Take a non-closed subspace $Y$ of $X$. Then (that is linear algebra) there is a complementary subspace $Z$ such that $X = Y \oplus Z$.

But caution ! In the context of Banach spaces some people write $X = Y \oplus Z$ with the meaning that $Y$ and $Z$ are topological complements, hence both are closed.

Conclusion: read definition carefully !

Answer 2

The following version is true: suppose $X=Y\oplus Z$ where $Y$ and $Z$ are linear subspaces which are orthogonal to each other. Then these subspaces are necessarily closed. Proof: let $\{y_n\} \subset Y$ and $y_n \to u$. Write $u=y_0+z_0$ where $y_o \in Y$ and $z_0 \in Z$. Then $y_n -y_0 \to z_0$. Now $\langle z_0, z_0 \rangle>=\lim \langle {y_n-y_0}, z_0 \rangle =0$ since $y_n-y_0$ is orthogonal to $z_0$. Hence $z-0=0$ and $u=y_0 \in Y$. Similarly $Z$ is also closed.