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I am currently reading a proof of the Hyperplane separation theorem. It uses the following:

If $X\subseteq\mathbb R^n$ is a convex set containing $0$, and $\alpha\in (0,1)$, then the closure of $\alpha X:=\{\alpha x\mid x\in X\}$ is contained in $X$.

This seems highly intuitive. However, I am unable to verify it formally. Any hint on how it can be proven?

Zuy
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  • This is probably not a terribly elegant argument, but my first inclination is to think "If $x \in \alpha X$, then $y= \frac{1}{\alpha}x \in X$, and $x$ is on the segment through $0$ and $y$. Apply convexity." – Xander Henderson Oct 04 '22 at 16:19
  • @XanderHenderson The problem is we're talking about the closure of $\alpha X$. – Zuy Oct 04 '22 at 16:26

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I don't think this is true without further conditions on $X$. Suppose: $$X = \big\{(x,y) \in \mathbb{R}^2 : y > 0\big\} \cup \{(0,0)\},$$ i.e. that $X$ consists of everything strictly above the $x$-axis in the plane, union the origin. $X$ is convex, but for any $\alpha > 0$ the closure of $\alpha X$ is everything weakly above the $x$-axis, which is not contained in $X$.

Pete Caradonna
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