Question: Let If $P\subseteq \mathbb{R}^n$ be a convex set. show that $int(P)$ is a convex set.

I know that a point $x$ is said to be the interior point of the set $P$ if there is an open ball centred at $x$ that is contained entirely in $P$. The set of all interior points of $P$ is denoted by $int(P)$.

Also a set is convex if $x,y \in P$ $\implies$ $(tx+(1-t)y) \in P$ for all $t \in (0,1)$.

How to go about the above proof?

  • 3
    $\begingroup$ Hint: if $U$ and $V$ are open, then $(1-t)U+tV$ is open for every fixed $t\in [0,1]$, as the continuous image of $U\times V$ under $(x,y)\longmapsto (1-t)x+ty$. Now if $U$ and $V$ are contained in $P$, then... $\endgroup$ – Julien Jun 23 '13 at 18:33

Assume $U:=int(P)$ is not convex. There are $x_0,y_0\in U$ and $t_0\in (0,1)$ such that $z_0:=t_0x_0+(1-t_0)y_0\notin U$. Keep the segment $x_0y_0$ with fixed length and fixed at, say, $x_0$. Then $y$ is a continuous function of $z$, when you move $z$ and keep the segment $x_0y$ with the same length as $x_0y_0$. (These constrains are not really important, we are just fixing the ideas) Therefore, given a neighborhood $V\subset U$ of $y_0$ there is a neighborhood $W$ of $z_0$ such that for all points in $W$ the corresponding $y$ is in $V$. Since $z_0\notin int(P)$ there are points, say $z_1$, in $W$ that do not belong to $P$. Let $y_1$ be the corresponding $y$ for that $z_1$. Then $x_0$ and $y_1$ are two points of $P$ such that there is a point, $z_1$, in the segment $x_0y_1$ that is not in $P$. This is a contradiction because $P$ was convex. Therefore the non-interior point $z_0$ doesn't exist.

To prove the continuity of the correspondence above use triangle inequality.


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