Algebraic Closure (convex Analysis)

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 $\operatorname A$ or $\operatorname_X A$.

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

The set ''A'' is algebraically closed if $A = \operatorname A$.
The set $\operatorname A \setminus \operatorname A$ is the algebraic boundary of ''A'' in ''X''.

Examples

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

If $A = \ \subseteq \R^2$ then
$0 \in (\operatorname \operatorname A) \setminus \operatorname A$. In particular, the algebraic closure need not be algebraically closed.
Here, $\overline=\operatorname \operatorname A = \ = (\operatorname A)\cup\$.

However, $\operatorname A =\overline$ 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

References

Bibliography

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{{Convex analysis and variational analysis

Convex analysis
Functional analysis
Mathematical analysis
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