Algebraic closure (convex analysis)

For the algebraic closure of a field, see Algebraic closure.

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 $\operatorname {acl} A$ or $\operatorname {acl} _{X}A$.

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

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

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

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

See also

• Algebraic interior

Bibliography

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