Questions regarding internal and interior points for a convex subset of a topological vector space

Question

Suppose that $X$ is a topological vector space, with a convex subset $A$. How do we show that if the vector $u$ is in the interior of $A$, then $u$ is an internal point of $A$ and if the interior of $A$ is nonempty with $u$ an internal point of $A$, then $u$ is in the interior of $A$? Finally, is there some good example of a subset of $\mathbb{R}^2$ with an internal point which is not an interior point?

So far, I have read that $x_0$ is an internal point of a subset $Y$ of $X$, provided that for any $y \in X$, there exists $\delta > 0$ such that $x_0 + ty \in Y$ for $|t| < \delta$, but other than that, I'm not really sure how to approach this.

Answer 1

In a finite-dimensional topological vector space, interior points and internal points coincide for convex sets.

This is no longer true in infinite dimension.

Here is a reference for these two facts: http://books.google.fr/books?id=4hIq6ExH7NoC&pg=PA200&lpg=PA200&dq=internal+point+not+interior&source=bl&ots=p9qW_ql0JE&sig=EMEVsG-xYxCxhFybWFSYmPpZttw&hl=fr&sa=X&ei=n1YNUaPlIKPU0gHmm4CwDQ&ved=0CDYQ6AEwAQ#v=onepage&q=internal%20point%20not%20interior&f=false

Now let us find a counter-example in $\mathbb{R}^2$ when $A$ is not convex.

Consider the set $A$ consisting of the union of:

• the region delimited by the graphs of $\sqrt{x}$ and $-\sqrt{x}$ over $[0,+\infty)$

• the region delimited by the graphs of $\sqrt{-x}$ and $-\sqrt{-x}$ over $(-\infty,0]$

• the line $y$-axis $\{x=0\}$

Then convince yourself that $(0,0)$ is internal, but not in the interior of $A$.