1
$\begingroup$

Let $X$ be a closed convex subset of $\mathbb{R}^n$ with $intX\neq\emptyset$. Then $X=\overline{intX}$.

Here is the idea: observe that $X=\overline X=X\cup \partial X = intX \cup \partial X \cup I$, where $I$ is the set of all isolated points of $X$. But a convex set with more than one point has no isolated points, so $X = intX\cup \partial X=intX\cup \partial (intX)=\overline{intX}.$

In order for this to be complete we need to show that under our conditions $\partial X=\partial (intX)$, but since $\partial X\supset\partial(intX)$ is true for any set in any topologicatl space we only really need to prove $\partial X\subset\partial(intX)$. I am a little unsure of how to go about it.

Thanks.

  • 1
    Try taking a sequence of points in $X$ converging to a point in $\partial X$ and seeing if you can obtain a sequence of points in the interior converging to the same limit point. Convexity here will be key. – 2017-02-16

1 Answers 1

1

Notice that $int(X)\subset X\land X=\bar{X}\implies \overline{int(X)}\subset X$, so you need to prove that $\forall x\in X,\exists (x_n)\subset int(X):x_n\rightarrow x$, so take some $x_0\in int(X)\neq\emptyset$ and define $x_n:=x_0+\frac{1}{n}\left( x-x_0\right)=\frac{1}{n}x+\left( 1-\frac{1}{n}\right)x_0$ where $x_n\in int(X)$ as $X$ convex $\implies int(X)$ convex and $y\in Y,z\in(\bar{Y})$ convex$\implies [y,z)\subset Y$.