calculus - Show that $(x, y] \subset \mbox{int}\; C$ where C is a convex set

Given $C \subset \mathbb{R}^n$ a convex set, $x \in \overline{C}$ and $y \in \mbox{int} \;C$. Show that $(x, y] \subset \mbox{int}\; C$.

I know the definition about convex set and I know that I need to prove for all point $p \in (x,y]$, there is a $\delta > 0$ such that $ p \in B(p,\delta) \subset C$. But I couldn't to prove this. I was thinking this a long time because this is apparently easy.

1 Answer

  1. Marks- Reply


    I have copied the following from some notes I had, so the notations are slightly different. I leave it to you to complete the argument using Theorem 2. (Drawing a picture will help in understanding what $\rho$ is).

    Theorem 1

    Let $C$ be a bounded convex set in $\mathbb R^{k}$ with non-empty interior. Let $x\in C^{0}.$ If $\left\Vert u\right\Vert =1$ then there is a unique number $\rho =\rho (u)$ such that $x+ru\in C$ if $% 0\leq r<\rho $ and $x+ru\notin C$ if $r>\rho .$

    Proof: let $\rho =\inf \{r>0:x+ru\notin C\}.$ It is easy to see that $\rho \in (0,\infty ).$ If $0\leq r<\rho $ then $x+ru\in C$ by definition. If $% r>\rho $ and $x+ru\in C$ then, for any $t\in \lbrack \rho ,r)$ we have $x+tu=% \frac{r-t}{r}x+\frac{t}{r}(x+ru)\in C$ by convexity. Thus, $\{r>0:x+ru\notin C\}$ does not intersect $[\rho ,r).$ We already know that it does not intersect $[0,\rho )$ and hence its infimum must be at least $r$ which means $\rho \geq r.$ This is a contradiction.

    Theorem 2

    If $C,x,u,\rho $ are as above then $x+ru\in C^{0}$ if $0\leq r<\rho $ and $% x+ru\notin C^{e}$ if $r>\rho $ (where $C^{e}$ is the exterior of $C).$

    Proof: let $0\leq r<\rho .$ There exists $\delta >0$ such that $B(x,\delta )\subset C.$ Let $r<s<\rho .$ If $0<\epsilon <(1-\frac{r}{s})\delta $ we claim that $B(x+ru,\epsilon )\subset C.$ If $y\in B(x+ru,\epsilon )$ then $% y=(1-\frac{r}{s})z+\frac{r}{s}(x+su)$ where $z$ is defined by this last equation. If we show that $z\in B(x,\delta )$ it would follow that $z$ and $% (x+su)$ both belong to $C$ and hence that $y\in C.$ This would complete the proof of the claim. We have $\left\Vert z-x\right\Vert =\left\Vert \frac{y-% \frac{r}{s}(x+su)}{(1-\frac{r}{s})}-x\right\Vert $

    $=\left\Vert \frac{y-\frac{r}{s}(x+su)-(1-\frac{r}{s})x}{(1-\frac{r}{s})}% \right\Vert =\left\Vert \frac{y-ru-x}{(1-\frac{r}{s})}\right\Vert <\frac{% \epsilon }{(1-\frac{r}{s})}<\delta .$ We have proved the first part of the theorem.

    Now let $r>\rho .$ Let $s\in (\rho ,r)$ and $\delta $ be as before. Let $% \beta =\frac{s}{r}.$ Let $0<\eta <\frac{r-s}{r}\delta .$ We claim that $% B(x+ru,\eta )\subset C^{c}$ and this would finish the proof. Let $v\in B(x+ru,\eta )$ and suppose, if possible, $v\in C.$ Define $w$ by $x+su=\beta v+(1-\beta )w.$ This would lead to the contradiction [that $x+su\in C$ (even though $s>\rho )]$ if we show that $w\in B(x,\delta ).$ However, $\left\Vert \frac{x+su-\beta v}{1-\beta }-x\right\Vert =\left\Vert \frac{x+su-\beta v-(1-\beta )x}{1-\beta }\right\Vert =\left\Vert \frac{\beta x+su-\beta v}{% 1-\beta }\right\Vert =\frac{\beta }{1-\beta }\left\Vert x+\frac{s}{\beta }% u-v\right\Vert =\frac{\beta }{1-\beta }\left\Vert x+ru-v\right\Vert <\eta \frac{\beta }{1-\beta }<\delta .$ The proof is now complete.

    Take $u$ to be $\frac {y-x} {\|y-x\|}$ in theorem 2. (The roles of $x$ and $y$ are switched).

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