How can I show that $\{x\in X: \langle x, x_0\rangle =0\}$ is closed in X?

Question

Let $X$ be a Hilbert space over $\Bbb{R}$ and fix $x_0\in H\setminus \{0\}$. Define $Q:=\{x\in X: \langle x, x_0\rangle =0\}\subset X$. I want to show that it is closed.

My idea was the following:

Let me define $F:X\rightarrow \Bbb{R};~~x\mapsto \langle x,x_0\rangle$. Then $F^{-1}(\{0\})=Q$. Furthermore let $(x_n)_n\subset X$ be a sequence s.t. $x_n\rightarrow x$. Then $$||F(x_n)-F(x)||=||\langle x_n-x,x_0\rangle||\leq ||x_n-x||\cdot ||x_0||\rightarrow 0$$ So $F$ is continuous. But since $\{0\}$ is closed in $X$ we deduce that $Q$ is closed in $X$.

Does this work?

Answer 1

This solution is almost correct. However, we actually have $F : X \to \mathbb{R}$, not $F : X \to X$.

You also technically showed that $F$ is sequentially continuous, not that $F$ is continuous. However, since we’re working over metric spaces (and presumably we are using the axiom of countable choice or stronger), the fact that $F$ is sequentially continuous implies that it is continuous. So this is still an acceptable proof.

Another approach would be noting that the map $(x, y) \mapsto \langle x, y \rangle : X^2 \to \mathbb{R}$ is continuous (which can easily be proved by a $\delta-\epsilon$ argument, and is a fundamental fact of inner product spaces). From there, it’s trivial to note that $F$ is continuous.

Answer 2

Yes, your approach is correct. By showing that the function $F: X → X; x → ⟨x, x_0⟩$ is continuous, and that the pre-image of $0$ under $F$ is $Q$, you have shown that $Q$ is closed.

You defined the function $F: X → X; x → ⟨x, x_0⟩$ You showed that $F^{-1}(\{0\}) = Q$, which means the set of all $x$ in $X$ such that $F(x) = 0$ is $Q$. You showed that if $(x_n)_n$ is a sequence in $X$ converging to $x$, then

$$||F(x_n) − F(x)|| = ||⟨x_n − x, x_0⟩|| ≤ ||xn − x||⋅||x_0|| → 0$$

This means that $F$ is continuous. Since $\{0\}$ is closed in $X$, and $F^{-1}(\{0\}) = Q$, it follows that $Q$ is closed in $X$. So your proof is correct, and it shows that $Q$ is closed in $X$.