Let $C \subset \mathbb{R}^{m}$ be a convex set. Prove that $\bar{C}$ is also convex
My attempt
Let $x,y \in \bar{C}$, this implies two things:
$x=\lim_{n\rightarrow ∞} x_{n}$ where $(x_{n})_{n \in \mathbb{N}}$ in $C$ and thus $x_{n} \in \bar{C}$
$y=\lim_{n\rightarrow ∞} x_{n}$ where $(y_{n})_{n \in \mathbb{N}}$ in $C$ and thus $y_{n} \in \bar{C}$
We have to prove that for all $t \in [0,1]$, $tx_{n}+(1-t)y_{n} \in C$, and this is $\lim_{n\rightarrow ∞} z_{n}=z$ where $z_{n}=tx_{n}+(1-t)y_{n}$ and $z=tx+(1-t)y$
Let's take $\lim_{n\rightarrow ∞} x_{n}=x$ and $\epsilon=\frac{\epsilon}{2t}$, then we have (by definition of limit), $\exists N_{1}\in \mathbb{N}$ such that $||x_{n}-x||<\frac{\epsilon}{2t}$ for every $n \geq N_{1}$
Then, let's take $\lim_{n\rightarrow ∞} y_{n}=y$ and $\epsilon=\frac{\epsilon}{2(1-t)}$, then we have (by definition of limit), $\exists N_{2}\in \mathbb{N}$ such that $||y_{n}-y||<\frac{\epsilon}{2(1-t)}$ for every $n \geq N_{2}$
Taking $N=max(N_{1},N_{2})$ we have:
$||tx_{n}+(1-t)y_{n}-(tx+(1-t)y)|| \leq ||t(x_{n}-x)+(1-t)(y_{n}-y)|| \leq t||x_{n}-x||+(1-t)||y_{n}-y|| < t\frac{\epsilon}{2t} + (1-t)\frac{\epsilon}{2(1-t)}=\epsilon $. And that is what we wanted.
P.S. I don't know if the epsilon thing is totally right because $t$ may be $0$ o $1$ and in that case epsilon doesn't exit.
