I need to prove that given $C$, a convex non-empty set:
$\forall y \in C \subseteq \mathbb{R}^n, \exists \epsilon>0$ | $x + \epsilon y \in C \subseteq \mathbb{R}^n \implies x \in int C$
Given the definition of interior of a set($\exists r > 0 | B(x,r) \subseteq C \implies x \in intC$), i know that i need to find a ball of radius $r(\epsilon)$, but i'm not being able to write an argument.
Thanks in advance.
Let $X$ be a topological vector space (over $\Bbb R$) containing the set $C$. The claim which you need to prove in general is not true. First, you should have “$\forall y\in X$” instead “$\forall y\in C$” (a counterexample is $X=\Bbb R^2$, $C=\Bbb R\times \{0\}$, $x=(0,0)$). Next, very probably $X$ should be finitely dimensional (because for an infinitely dimensional topological vector space the the claim seems to be not true. Maybe even it fails for each infinitely dimensional topological vector space). But if $X$ is finitely dimensional then it is topologically isomorphic to the space $\Bbb R^n$ endowed with the standard norm. So without loss of generality we may suppose that $X=\Bbb R^n$. Let $e_1,\dots, e_n$ be the standard orts of the space $R^n$. Put $E=\{e_1,\dots,e_n,-e_1,\dots,-e_n\}$. Since the set $E$ is finite then there exists $\epsilon>0 $ such that $x+\varepsilon y\in C$ for each $y\in E$. Since the set $C$ is convex, $x+\epsilon\operatorname{conv} E\subset C$. Using the inequality between the arithmetic and quadratic means, we can easily show that $\operatorname{conv} E\supset B(0,\epsilon/\sqrt{n})$. Thus $x+B(0,\epsilon/\sqrt{n})\subset C$, so $x\in\operatorname{int} C$.