How to prove the following assertion about Hilbert spaces.

Question

Let $X$ be a separable Hilbert space and $Y$ a closed subset. Then I want to conclude that $Y$ is separable. So I wanted to invoke the theorem that says that a Hilbert space is separable iff it has a complete orthonormal system, but then I realized that since $Y \subset X$ then the same complete orthonormal system works for $Y$.

So I think there is something wrong in my reasoning, because the above argument seems a little bit awful for me. Then Can someone help me to prove that theorem please?

Thanks a lot in advance.

Answer 1

Hint: Since $Y$ is closed, the orthogonal projection $P_Y$ is well defined. Now take a countable dense subset of $X$ and consider its projection onto $Y$.

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More specifically: let $A$ be a countable dense subset in $X$. Take an arbitrary open set $U$ in $Y$, and consider $P_Y^{-1}(U) \subset X$, which is open by continuity. Now, by denseness of $A$ there exists an element $a \in A$ such that $a \in P_Y^{-1}(U)$. Then $P_Y(a) \in U$, so $P_Y(A)$ is the desired countable dense subset (every open set in $Y$ contains an element of the dense subset).

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Note that this also proves that if you have $f\colon H \to H'$ continuous and surjective, and $H$ is separable, then $H'$ is separable.

Answer 2

Seems to me that any proof involving concepts specific to Hilbert space is "wrong", simply because this is true in any metric space!

Say $X$ is a separable metric space and $Y\subset X$. Then $Y$ is separable:

Say $(x_n)$ is a countable dense subset of $X$. Let $S$ be the set of pairs $(n,j)$ such that there exists $y\in Y$ with $d(y,x_n)<1/j$. For each $(n,j)\in S$ choose $y_{n,j}\in Y$ with $d(y_{n,j},x_n)<1/j$. Then the set of all $y_{n,j}$ for $(n,j)$ in $S$ is a countable subset of $Y$. And it's dense in $Y$:

Say $y\in Y$ and $j$ is a positive integer. Since $(x_n)$ is dense in $X$ there exists $n$ such that $d(y,x_n)<1/j$. Hence $(n,j)\in S$, so $d(y_{n,j},x_n)<1/j$. So $d(y,y_{n,j})<2/j$.