Limit points

# Limit points

## Idea

Given a space $S$, a subspace $A$ of $S$, and a concrete point $x$ in $S$, $x$ is a limit point of $A$ if $x$ can be approximated by the contents of $A$.

There are several variations on this idea, and the term ‘limit point’ itself is ambiguous (sometimes meaning Definition , sometimes Definition .

## Definitions

The classical definitions apply when $S$ is a topological space. Then $A$ may be thought of as a subset of (the underlying set of) $S$, and $x$ as an element. In order to apply the definitions in constructive mathematics, there needs to be an inequality relation $\ne$ on the points of $S$; in classical mathematics, this is taken to be the denial inequality, as usual. (We need not assume that $\ne$ is an apartness relation nor any compatibility between $\ne$ and the topology, at least for the definitions; although it's quite possible that some classical theorems will require such assumptions.)

For the most general definitions, let $\kappa$ be a collection of cardinal numbers. (We might want $\kappa$ to have some closure properties akin to those of an arity class, but the definition there is not quite what we want.) Recall that a $\kappa$-ary? indexed subset of $S$ is a function $B\colon I \to S$ such that the cardinality of $I$ belongs to the class $\kappa$; a point $y$ is in $B$ (as an indexed subset) if $y$ belongs to the range of $B$ (as a function), and $y$ is out of $B$ if $y$ is inequal ($\ne$) to every point in $B$.

###### Definitions

The point $x$ is a $\kappa$-adherent point of the subspace $A$ if, for each neighbourhood $U$ of $x$, for each $\kappa$-ary indexed subset $B$ of the intersection $U \cap A$, there is an element $y$ of $U \cap A$ that is out of $B$. Slightly more strongly, $x$ is a $\kappa$-accumulation point (or $\kappa$-cluster point) of $A$ if, for each neighbourhood $U$ of $x$, for each $\kappa$-ary indexed subset $B$ of $U \cap A$, there is an element $y \ne x$ of $U \cap A$ that is out of $B$. (Alternatively, take $U$ to be a punctured neighborhood?, but that won't work constructively in general.)

Every $\kappa$-accumulation point is a $\kappa$-adherent point; the converse holds if every $k \in \kappa$ satisfies $k + 1 \in \kappa$ (and then one usually says ‘accumulation’ rather than ‘adherent’). Also, if $\kappa \subseteq \lambda$, then every $\lambda$-(adherent/accumulation) point is a $\kappa$-(adherent/accumulation) point.

It immediately follows that the following classical special cases are in order of increasing strength:

• Using $\kappa = 1 = \{0\}$:

###### Definition

The point $x$ is an adherent point of the subspace $A$ if, for every neighbourhood $U$ of $x$, the intersection $U \cap A$ is inhabited (nonempty).

• Using $\kappa = 1$ again:

###### Definition

The point $x$ is an accumulation point of the subspace $A$ if, for every punctured neighbourhood $U$ of $x$, the intersection $U \cap A$ is inhabited.

• Using $\kappa = \omega = \{0, 1, 2, \ldots\}$:

###### Definition

The point $x$ is an $\omega$-accumulation point of the subspace $A$ if, for every neighbourhood $U$ of $x$, the intersection $U \cap A$ is infinite.

• Using $\kappa = \omega_1 = \{0, 1, 2, \ldots, \aleph_0\}$:

###### Definition

The point $x$ is a condensation point of the subspace $A$ if, for every neighbourhood $U$ of $x$, the intersection $U \cap A$ is uncountable.

## Properties

The subspace $A$ is closed iff every adherent point of $A$ belongs to $A$ and iff every accumulation point of $A$ belongs to $A$. (Thus one may say that $A$ is closed iff every limit point of $A$ belongs to $A$ without ambiguity.)

More generally, the closure of $A$ is the set of all adherent points of $A$. Classically (using excluded middle, or more generally if $S$ has decidable equality), the closure of $A$ is the union of $A$ and its set of accumulation points.

Classically, a point in $A$ that is not an accumulation point of $A$ is precisely an isolated point?. (Constructively, each of these is stronger than the negation of the other.)

A justification for the terminology ‘limit point’ for an accumulation point is that the concept of limit of a function? approaching a point really only makes sense approaching an accumulation point. (This is for essentially the same reason that every function is continuous at an isolated point.) More precisely, every answer whatsoever satisfies the definition of $\lim_c f$ if $c$ is a non-accumulation point of $dom f$ (because the improper filter converges everywhere).