Adherent point

 ( General Topology ) 

In mathematics, an adherent point (also closure point or point of closure or contact point)Steen, p. 5; Lipschutz, p. 69; Adamson, p. 15. 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 $x\; \backslash in\; X$ is an adherent point for $A$ if and only if $x$ is in the closure of $A,$ thus

$x\; \backslash in\; \backslash operatorname\{Cl\}\_X\; A$ if and only if for all open subsets $U\; \backslash subseteq\; X,$ if $x\; \backslash in\; U\; \backslash text\{\; then\; \}\; U\; \backslash cap\; A\; \backslash neq\; \backslash varnothing.$

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.

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

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

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

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 $X.$

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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 $\backslash left(x\_n\backslash right)\_\{n\; \backslash in\; \backslash N\}$ be such a sequence and let $x$ be its limit. Then by definition of limit, for all neighbourhoods $U$ of $x$ there exists $n\; \backslash in\; \backslash N$ such that $x\_n\; \backslash in\; U$ for all $n\; \backslash geq\; N.$ In particular, $x\_N\; \backslash in\; U$ and also $x\_N\; \backslash in\; S,$ 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 $S\; =\; \backslash \{\; 0\; \backslash \}$ as a subset of $\backslash R.$ Then the only sequence in $S$ is the constant sequence $0,\; 0,\; \backslash ldots$ whose limit is $0,$ but $0$ is not a limit point of $S;$ it is only an adherent point of $S.$

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See also

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Notes

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Citations

• Adamson, Iain T., A General Topology Workbook, Birkhäuser Boston; 1st edition (November 29, 1995). .
• Apostol, Tom M., Mathematical Analysis, Addison Wesley Longman; second edition (1974).
• Lipschutz, Seymour; Schaum's Outline of General Topology, McGraw-Hill; 1st edition (June 1, 1968). .
• Lynn Steen, J.A.Seebach, Jr., Counterexamples in topology, (1970) Holt, Rinehart and Winston, Inc..

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