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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.

Boar
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nimeshkiranverma
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5 Answers5

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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.

Boar
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acyrl
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  • 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. – Jale'de jaled Jun 13 '20 at 00:23
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    @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... – PinkyWay Nov 19 '21 at 20:43
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    I agree that "Now taking the limit gives the correct result" is a misstatement. You can have a (non-convex) set, two points in the closure of that set, and sequences going to those two points where each segment between corresponding points in the sequences is in the set, but the segment between the limit points is not. Take for example, the punctured plane with the two limit points being a +1 and -1 on the x-axis. Let your sequences approach from the upper half plane. Clearly we need to use convexity more than just forcing the line segments between the sequence points are in the space. – Kyle Parsons Sep 08 '22 at 18:32
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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. $(z+W) \cap C \neq \emptyset$.

Seirios
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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.

Justin Young
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  • for future visitors, note that this answer depends on https://math.stackexchange.com/questions/543791/the-closure-of-a-product-is-the-product-of-closures and https://math.stackexchange.com/questions/114462/a-map-is-continuous-if-and-only-if-for-every-set-the-image-of-closure-is-contai – D.R. Mar 04 '22 at 17:50
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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, $$tx+(1-t)y=\lim_{k\to\infty}(tx_{k}+(1-t)y_{k})\in \bar A$$

Boar
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Tomás
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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.

HorizonsMaths
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