Algebraic closure (convex analysis)

Algebraic closure of a subset ${\displaystyle A}$of a vector space ${\displaystyle X}$is the set of all points that are linearly accessible from ${\displaystyle A}$. It is denoted by ${\displaystyle \operatorname {acl} A}$or ${\displaystyle \operatorname {acl} _{X}A}$.

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

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

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

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

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

However, ${\displaystyle \operatorname {acl} A={\overline {A}}}$for every finite-dimensional convex set A.

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

• Algebraic interior

• Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.