Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
Where are you stuck? –  Davide Giraudo Sep 25 '12 at 19:12

5 Answers 5

Pick two points in the closure of the set. Say $x,y \in \bar{X}$. So there exits 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.

share|improve this answer
Thanks man!!! great help.. –  nimesh.iitd Sep 25 '12 at 19:17

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, $$tx+(1-t)t=\lim (tx_{k}+(1-t)y_{k})\ \in Closure(A)$$

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

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. $(z+W) \cap C \neq \emptyset$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.