compact subset of infinite-dimensional inner product space has empty interior

Question

Could any one help me to show that every compact subset of infinite-dimensional inner product space $X$ has empty interior. Thanks in advance.

Answer 1

If $K$ is compact and if it has an interior point $x$ then $\overset {-}{B}(x,r) \subset K$ for some $r>0$. This close ball is then compact. The closed unit ball is teh image of this closed ball undet the homeomorphism $y \to \frac 1 r (y-x)$. Hence the closed unit ball is compact which implies that the space is finite dimensional.

Answer 2

You can prove it by contradiction. Let $K$ be the compact and suppose $x_0$ is an interior point. Then there exists a ball $B(x_0,r)\subset K$, with $r>0$. Let $e_1,e_2,\dots$ an infinite orthonormal system (we know there are infinitely many linearly independent vectors and we can orthonormalize them using the Hilbert-Schmidt orthogonalization procedure). Then the infinitely many different points (vectors) $x_0+\frac{r}{2}e_i$ are in $B(x_0,r)$ (hence in $K$) within a distance from each other $r\sqrt{2}/2$ by Pythagorean Theorem. As a consequence, your set $K$ does not admit, say, a finite $r\sqrt{2}/4$- net and is not compact (compact = totally bounded + complete in any metric space). Conclusion: your set does not have interior points.