Algebraic closure (convex analysis)

Template:No footnotes Template:One source Template:About Algebraic closure of a subset $A$ of a vector space $X$ is the set of all points that are linearly accessible from $A$. It is denoted by $\mathrm{acl}A$ or ${\mathrm{acl}}_{X}A$.

A point $x\in X$ is said to be linearly accessible from a subset $A\subseteq X$ if there exists some $a\in A$ such that the line segment $[a,x):=a+[0,1)(x-a)$ is contained in $A$.

Necessarily, $A\subseteq \mathrm{acl}A\subseteq \mathrm{acl}\mathrm{acl}A\subseteq \bar{A}$ (the last inclusion holds when X is equipped by any vector topology, Hausdorff or not).

The set A is algebraically closed if $A=\mathrm{acl}A$. The set $\mathrm{acl}A\setminus \mathrm{aint}A$ is the algebraic boundary of A in X.

Examples

The set $\mathbb{Q}$ of rational numbers is algebraically closed but ${\mathbb{Q}}^c$ is not algebraically open

If $A=\{(x,y)\in {ℝ}^2:0<y<x^2\}\subseteq {ℝ}^2$ then $0\in (\mathrm{acl}\mathrm{acl}A)\setminus \mathrm{acl}A$. In particular, the algebraic closure need not be algebraically closed. Here, $\bar{A}=\mathrm{acl}\mathrm{acl}A=\{(x,y)\in {ℝ}^2:0\le y\le x^2\}=(\mathrm{acl}A)\cup \{0\}$.

However, $\mathrm{acl}A=\bar{A}$ for every finite-dimensional convex set A.

Moreover, a convex set is algebraically closed if and only if its complement is algebraically open.

See also

• Algebraic interior

Bibliography

• Template:Narici Beckenstein Topological Vector Spaces