Convex sets in Linear topological spaces

A (real) linear topological space is a real linear space (vector space) $Ε$ with a Hausdorff topology such that:

II) scalar multiplication is continuous

For $x$ and $у$ in $E$, denote by $L(x, y)$ the set of all points $z$ such that $z= λ_1x +λ_2y$ with $0 \le λ_i \le 1$ and $λ_1 + λ_2 = 1$.

A subset $A$ of $E$ is convex iff whenever $x$ and $y$ belong to $A$,then $L(x,y) \subset A$

My Question:-

If $A$ is convex and $x \in A^{\mathrm{o}}$, $у \in \bar{A}$, show that $L(x, у)-{у} \subset А$

I Know that $A^{\mathrm{o}}$ and $\bar{A}$ are convex sets, but what should I do then ??

• Draw a little picture of the situation. If $x \in A^\circ$ then there is some open $U\subset A$ containing $x$. Now shift and scale (shrink) the set $U$ as you move from $x$ to $y$ 'linearly' so that the resulting set stays inside $A$. – copper.hat Oct 31 '17 at 2:54
• @copper.hat: that picture works well to show that $x\in \mathring{A}$, $y\in A$ implies $[x,y)\subset \mathring{A}$. Here $y \in \bar{A}$. – Orest Bucicovschi Oct 31 '17 at 3:16
• @orangeskid: It works for $y$ in the closure as well. – copper.hat Oct 31 '17 at 3:20
• In my opinion, this particular fact captures the essence of convexity. – copper.hat Oct 31 '17 at 3:21
• @copper.hat: I agree, the cone without vertex lies inside. $\mathring{A}$. Good call, also about the relative interior. – Orest Bucicovschi Oct 31 '17 at 3:33

In fact you can show that if $x\in \mathring{A}$, and $y\in \bar A$, then $[x,y)\in \mathring{A}$. Let $\lambda \in (0, 1]$. We want to show that $$\lambda x + (1-\lambda)y + \delta_{\lambda} \in A$$ if $\delta_{\lambda}$ is small enough. What we know is that in every neighborhood of $0$ there exists $\delta$ so that $y+\delta \in A$. So now we rewrite $$\lambda x + (1-\lambda)y + \delta_{\lambda}= (1-\lambda)(y + \delta) + \lambda( x- \frac{1-\lambda}{\lambda}\delta + \frac{1}{\lambda}{\delta_\lambda})$$ Recall that $\lambda \in (0,1]$ is fixed. If $\delta$ and $\delta_{\lambda}$ are small enough, $- \frac{1-\lambda}{\lambda}\delta + \frac{1}{\lambda}{\delta_\lambda}$ will lie in the neighborhood $W_{x}$ with the property that $x+W_{x}\subset A$.