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Today I was going through my text book exercise and got hold of the following question.
Let $X$ be a normed linear space with norm $\lVert\cdot\rVert$ and $A$ a nonempty convex subset of $X$. Then prove that $\operatorname{Closure}(A)$ is also a convex subset of $X$.
I gave it a try but could not succeed.
Pick two points in the closure of the set. Say $x,y \in \bar{A}$. So there exist some sequences $\{x_i\}$ and $\{y_i\}$ in $A$ which tend to $x$ and $y$ respectively.
But for all $i$ and $0 < t < 1$ we have that: $$ tx_{i} + (1-t)y_{i} \in A$$
Now taking limits gives the correct result.
In fact, the result is true in any topological vector space $X$.
Let $C$ be a nonempty convex subset of $X$. For $x,y \in \overline{C}$ and $\lambda \in [0,1]$, we prove that any neighborhood of $z= \lambda x+ (1-\lambda)y$ intersects $C$. So let $W$ be a neighborhood of $0$. Because $(u,v) \mapsto \lambda u+(1-\lambda)v$ is continuous, there exist open subsets $U$ and $V$ such that $\lambda U+(1-\lambda)V \subset W$; $x+U$ (resp. $y+V$) is an open neighborhood of $x$ (resp. of $y$) so there exists $x_1 \in (x+U) \cap C$ (resp. $y_1 \in (y+V) \cap C$). Therefore, $z_1= \lambda x_1+(1-\lambda)y_1 \in C$ because $C$ is convex and $z_1 \in \lambda (x+U)+(1-\lambda)(y+V) \subset z+W$, ie. $(z+W) \cap C \neq \emptyset$.
Perhaps to get away from sequences, for $t\in I$, consider the continuous map $t: X \times X\to X$ given by $(x,y)\mapsto tx + (1-t)y$. Since $A$ is convex, $t(A\times A) \subset A$.Thus, by continuity we have $$t\left (\overline A \times \overline A\right ) \subset \overline{t(A\times A)} \subset \overline A,$$ therefore $\overline A$ is convex.
Let $x,y\in \bar A$, $t\in [0,1]$. Choose sequences $x_{k},y_{k}\in A$ such that $x_{k}\rightarrow\ x$ and $y_{k}\rightarrow\ y$. Because $A$ is convex, we have $tx_{k}+(1-t)y_{k}\in A$. Then, $$tx+(1-t)y=\lim_{k\to\infty}(tx_{k}+(1-t)y_{k})\in \bar A$$
Given $x,y \in \overline{A}$, there exist sequences $(x_n)_n, (y_n)_n \subset A$ such that $\lim_nx_n=x$ and $\lim_ny_n=y$. Since $A$ is convex, and $x_n,y_n \in A$ for every $n$, we have $tx_n+(1-t)y_n \in A$ for every $t \in [0,1]$. It follows that $tx+(1-t)y=\lim_n[tx_n+(1-t)y_n] \in \overline{A}$. Hence $\overline{A}$ is convex.
