Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to show that if $X$ is a pre-Hilbert space and $A$ is a subset of $X$ with an nonempty interior, then $A^{\perp} = \{ 0 \}$.

I tried to assume the contrary, then there would be an $x \ne 0$ from $A^{\perp}$ such that $$ \langle x, a \rangle = 0 $$ for some interior point of $A$. Because $a$ is an interior point, there would be an open ball $B_r(a)$ around $a$ with $B_r(a) \subseteq A$.

I know the following facts,

a) $A^{\perp}$ is closed

b) $A$, $A^{\perp}$ are linear, i.e. they are infinite

c) $A \cap A^{\perp} = \{ 0 \}$

d) $(A^{\perp})^{\perp} \subseteq \overline{A}$

e) the scalar product and the norm on $X$ are continous

but I am unable to employ them in any useful way?

share|cite|improve this question
up vote 1 down vote accepted

Hint: the orthogonal of $A$ is the same thing as the orthogonal of the vector sub-space generated by $A$ (use finite sums to see that). What about a vector sub-space in a normed space which has a non-empty interior?

share|cite|improve this answer
... then in equals the whole space, okay now its simple :) – Stefan Nov 29 '12 at 18:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.