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

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?

• $x\mapsto\langle x,x_0\rangle$ is a continuous function. As $\Bbb R$ is Hausdorff, any fibre (in particular at zero) is closed Jan 16 at 19:31
• @FShrike so my proof is correct, thanks! Jan 16 at 19:32
• Your proof is correct, yes, barring the typo $F:X\to X$ Jan 16 at 19:34
• Is there any way to prove the statement using that $F$ is a closed bounded linear transformation? Jan 16 at 19:53

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.

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 $$X$$.

• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Jan 16 at 20:26