Adherent point

In mathematics, an adherent point (also closure point or point of closure or contact point) of a subset $$A$$ of a topological space $$X,$$ is a point $$x$$ in $$X$$ such that every neighbourhood of $$x$$ (or equivalently, every open neighborhood of $$x$$) contains at least one point of $$A.$$ A point $$x \in X$$ is an adherent point for $$A$$ if and only if $$x$$ is in the closure of $$A,$$ thus
 * $$x \in \operatorname{Cl}_X A$$ if and only if for all open subsets $$U \subseteq X,$$ if $$x \in U \text{ then } U \cap A \neq \varnothing.$$

This definition differs from that of a limit point of a set, in that for a limit point it is required that every neighborhood of $$x$$ contains at least one point of $$A$$ $$x.$$ Thus every limit point is an adherent point, but the converse is not true. An adherent point of $$A$$ is either a limit point of $$A$$ or an element of $$A$$ (or both). An adherent point which is not a limit point is an isolated point.

Intuitively, having an open set $$A$$ defined as the area within (but not including) some boundary, the adherent points of $$A$$ are those of $$A$$ including the boundary.

Examples and sufficient conditions
If $$S$$ is a non-empty subset of $$\R$$ which is bounded above, then the supremum $$\sup S$$ is adherent to $$S.$$ In the interval $$(a, b],$$ $$a$$ is an adherent point that is not in the interval, with usual topology of $$\R.$$

A subset $$S$$ of a metric space $$M$$ contains all of its adherent points if and only if $$S$$ is (sequentially) closed in $$M.$$

Adherent points and subspaces
Suppose $$x \in X$$ and $$S \subseteq X \subseteq Y,$$ where $$X$$ is a topological subspace of $$Y$$ (that is, $$X$$ is endowed with the subspace topology induced on it by $$Y$$). Then $$x$$ is an adherent point of $$S$$ in $$X$$ if and only if $$x$$ is an adherent point of $$S$$ in $$Y.$$

By assumption, $$S \subseteq X \subseteq Y$$ and $$x \in X.$$ Assuming that $$x \in \operatorname{Cl}_X S,$$ let $$V$$ be a neighborhood of $$x$$ in $$Y$$ so that $$x \in \operatorname{Cl}_Y S$$ will follow once it is shown that $$V \cap S \neq \varnothing.$$ The set $$U := V \cap X$$ is a neighborhood of $$x$$ in $$X$$ (by definition of the subspace topology) so that $$x \in \operatorname{Cl}_X S$$ implies that $$\varnothing \neq U \cap S.$$ Thus $$\varnothing \neq U \cap S = (V \cap X) \cap S \subseteq V \cap S,$$ as desired. For the converse, assume that $$x \in \operatorname{Cl}_Y S$$ and let $$U$$ be a neighborhood of $$x$$ in $$X$$ so that $$x \in \operatorname{Cl}_X S$$ will follow once it is shown that $$U \cap S \neq \varnothing.$$ By definition of the subspace topology, there exists a neighborhood $$V$$ of $$x$$ in $$Y$$ such that $$U = V \cap X.$$ Now $$x \in \operatorname{Cl}_Y S$$ implies that $$\varnothing \neq V \cap S.$$ From $$S \subseteq X$$ it follows that $$S = X \cap S$$ and so $$\varnothing \neq V \cap S = V \cap (X \cap S) = (V \cap X) \cap S = U \cap S,$$ as desired. $$\blacksquare$$

Consequently, $$x$$ is an adherent point of $$S$$ in $$X$$ if and only if this is true of $$x$$ in every (or alternatively, in some) topological superspace of $$X.$$

Adherent points and sequences
If $$S$$ is a subset of a topological space then the limit of a convergent sequence in $$S$$ does not necessarily belong to $$S,$$ however it is always an adherent point of $$S.$$ Let $$\left(x_n\right)_{n \in \N}$$ be such a sequence and let $$x$$ be its limit. Then by definition of limit, for all neighbourhoods $$U$$ of $$x$$ there exists $$n \in \N$$ such that $$x_n \in U$$ for all $$n \geq N.$$ In particular, $$x_N \in U$$ and also $$x_N \in S,$$ so $$x$$ is an adherent point of $$S.$$ In contrast to the previous example, the limit of a convergent sequence in $$S$$ is not necessarily a limit point of $$S$$; for example consider $$S = \{ 0 \}$$ as a subset of $$\R.$$ Then the only sequence in $$S$$ is the constant sequence $$0, 0, \ldots$$ whose limit is $$0,$$ but $$0$$ is not a limit point of $$S;$$ it is only an adherent point of $$S.$$