Convex set also has convex interior (Corollary)

Question

While studying convex analysis, i came across two statements:

First statement:

Let $S$ be a convex set in $R^{n}$ with a nonempty interior. Let $x_{1} \in \operatorname{cl} S$ and $x_{2} \in$ int
S. Then $\lambda x_{1}+(1-\lambda) x_{2} \in$ int $S$ for each $\lambda \in(0,1)$. I understood the proof regarding this one.

Corollary:

Let $S$ be a convex set. Then int $S$ is convex.

My attempt:

For int $S$ to be convex, one has to let $x_{1}$, $x_{2}$ $\in$ int $S$ and show that $\lambda x_{1}+(1-\lambda) x_{2} \in int(S)$. I see that is almost what the first statement says. However, in statement number one, $x_{1}$ belongs to the closure of the set $S$ and i need it in $int (S)$ to conclude.

In an attempt to solve this issue, i was trying to use set inclusions such as $S \subset$ $cl(S)$ or $cl(S) = S \cup \partial S$ to conclude that $x_{1} \in cl(S)$ $\implies$ $x_{1} \in int (S)$ with the help of the first statement.

However, i just can't figure out this corollary.

Can someone help?

Thanks in advance, Lucas

Answer 1

You cannot conclude $x_{1} \in \text{cl}(S)$ $\implies$ $x_{1} \in \text{int} (S)$ because it is wrong. Just observe that the statement

Let $x_{1} \in \operatorname{cl} S$ and $x_{2} \in$ int S. Then $\lambda x_{1}+(1-\lambda) x_{2} \in \text{int} (S)$ for each $\lambda \in(0,1)$

is more general than

Let $x_{1}$, $x_{2}$ $\in$ int $S$. Then $\lambda x_{1}+(1-\lambda) x_{2} \in \text{int} (S)$ for each $\lambda \in(0,1)$.

In fact, if $x_{1}, x_{2} \in$ int $S$, then $x_1 \in \text{int} (S) \subset S \subset \text{cl}(S)$ and $x_2 \in \text{int} (S)$.