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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.

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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.

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 $A$.

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@Libertron Did this help? – 1015 Jun 1 '13 at 12:00

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