Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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?

share|cite|improve this question
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... – 1015 Jun 23 '13 at 18:33

Assume $U:=int(P)$ is not convex. There 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 $z_0$ is not interior there are points, say $z_1$, in $V$ 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.

share|cite|improve this answer

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.