# Is closure of convex subset of $X$ is again a convex subset of $X$?

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 $A$ is a non empty 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 $x,y \in \bar{X}$. So there exist some sequences $\{x_i\}$ and $\{y_i\}$ in $X$ which tend to $x$ and $y$ respectively.

But for all $i$ and $0 < t < 1$ we have that: $$tx_{i} + (1-t)y_{i} \in X$$

Now taking limits gives the correct result.

• the proof is a little weak, first need to show really there exists $x_i, y_i$ completely in $X$ which have limits $x,y\in \bar X$ and "Now taking limits gives the correct result" must be justified. Jun 13, 2020 at 0:23
• @Jale'dejaled, that is a proposition. $x\in\overline A\iff$ there is a sequence $(a_k)_k\in A$ converging to $A$. The assumption is that $X$ is nonempty, so... Nov 19, 2021 at 20:43

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 $$x,y\in Closure(A)$$, $$t\in [0,1]$$. Choose sequences $$x_{k},y_{k}\in D$$ such that $$x_{k}\rightarrow\ x$$ and $$y_{k}\rightarrow\ y$$. Because A is convex, we have $$tx_{k}+(1-t)y_{k}\in D$$. Then, $$tx+(1-t)t=\lim (tx_{k}+(1-t)y_{k})\ \in \bar A$$

• Please what about the converse? That is is $\overline{A}$ convex, is $A$ convex? Thanks a lot. Jun 23, 2020 at 17:16
• No and there is a simple example in $\mathbb{R}$. Jun 25, 2020 at 11:30

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.

• How is this different from the answer with the most upvotes? Jul 22, 2019 at 22:53