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. ==See also==