§ 7. Limits · General Topology

7. Limits

1. Limit of a filter

Definition 1

Let $X$ be a topological space and $\mathfrak{F}$ a filter on $X$ . A point $x \in X$ is said to be a limit point (or simply a limit) of $\mathfrak{F}$ , if $\mathfrak{F}$ is finer than the neighbourhood filter $\mathfrak{B}(x)$ of $x$ ; $\mathfrak{F}$ is also said to converge (or to be convergent) to $x$ . The point $x$ is said to be a limit of a filter base $\mathfrak{B}$ on $X$ , and $\mathfrak{B}$ is said to converge to $x$ , if the filter whose base is $\mathfrak{B}$ converges to $x$ .

This definition, together with Proposition 4 of § 6, no. 3, gives the following criterion:

Proposition 1

A filter base* $\mathfrak{B}$ on a topological space $X$ converges to $x$ if and only if every set of a fundamental system of neighbourhoods of $x$ contains a set of $\mathfrak{B}$ .

In accordance with the terminology introduced in § 1, no. 2 we can state Proposition 1 in the following way: $\mathfrak{B}$ converges to $x$ if and only if there are sets of $\mathfrak{B}$ as near as we please to $x$ .

If a filter $\mathfrak{F}$ converges to $x$ , then every filter finer than $\mathfrak{F}$ also converges to $x$ , by reason of Definition 1. Likewise, if the topology of $X$ is replaced by a coarser topology, the neighbourhood filter of $x$ is replaced by a coarser filter ( $\S$ 2, no. 2, Proposition 3), and therefore $\mathfrak{F}$ still converges to $x$ in this new topology.

We can therefore say that the finer the topology, the fewer convergent filters there are in this topology. In particular, in the discrete topology, the only convergent filters are the neighbourhood filters, for these are the trivial ultrafilters on $X$ ( $\S$ 6, no. 4).

Let $\Phi$ be a set of filters on $X$ , all of which converge to the same point $x$ ; the neighbourhood filter $\mathfrak{B}(x)$ is coarser than all the filters of $\Phi$ , hence also coarser than the intersection $\mathfrak{J}$ of these filters; in other words, $\mathfrak{J}$ also converges to $x$ .

Proposition 2

A filter* $\mathfrak{F}$ on a topological space $X$ converges to a point $x$ if and only if every ultrafilter which is finer than $\mathfrak{F}$ converges to $x$ .

This is an immediate consequence of the preceding remarks and Proposition 7 of § 6, no. 4.

In general a filter can have several distinct limit points; we shall revert to this question in § 8, no. 1.

2. Cluster point of a filter base

Definition 2

In a topological space* $X$ , a point $x$ is a cluster point of a filter base $\mathfrak{B}$ on $X$ if it lies in the closure of all the sets of $\mathfrak{B}$ .

If $x$ is a cluster point of a filter base $\mathfrak{B}$ , it is also a cluster point of every equivalent filter base by reason of § 6, no. 3, corollary to Proposition 4; in particular, $x$ is a cluster point of the filter whose base is $\mathfrak{B}$ .

Proposition 3

A point x is a cluster point of a filter base B if and only if every set of a fundamental system of neighbourhoods of x meets every set of B.

This follows immediately from the definitions.

This proposition and Corollary 2 to Proposition 1 of § 6, no. 2 show that the property "x is a cluster point of the filter $\mathfrak{F}$ " is equivalent to the property "there is a filter which is finer than both $\mathfrak{F}$ and the neighbourhood filter of x". In other words:

Proposition 4

A point x is a cluster point of a filter $\mathfrak{F}$ if and only if there is a filter finer than $\mathfrak{F}$ which converges to x.

In particular, every limit point of a filter $\mathfrak{F}$ is a cluster point of $\mathfrak{F}$ .

Corollary

An ultrafilter $\mathfrak{U}$ converges to a point x if and only if x is a cluster point of $\mathfrak{U}$ .

If x is a cluster point of a filter $\mathfrak{F}$ , it is also a cluster point of every filter coarser than $\mathfrak{F}$ ; likewise, if we replace the topology of X by a coarser topology, x remains a cluster point of $\mathfrak{F}$ in the new topology.

The set of cluster points of a filter base $\mathfrak{B}$ on X is by definition the set $\bigcap_{M \in \mathfrak{B}} \overline{M}$ , whence

Proposition 5

The set of cluster points of a filter base on a topological space X is closed in X.

Proposition 6

Let $\mathfrak{B}$ be a filter base on a subset A of a topological space X. Then every cluster point of $\mathfrak{B}$ in X belongs to $\overline{A}$ ; and conversely every point of $\overline{A}$ is a limit point of a filter on A.

The first assertion is trivial; on the other hand, if $x \in \overline{A}$ , the trace on A of the neighbourhood filter of x in X is a filter on A which evidently converges to x.

Remark

A filter on a topological space need have no cluster points (and a fortiori no limit points); for example, in an infinite discrete space the filter of complements of finite subsets has no cluster points. Spaces in which every filter has a cluster point play an important role in mathematics, and we shall study them in § 9.

3. Limit point and cluster point of a function

Definition 3

Let f be a mapping of a set X into a topological space Y, and let $\mathfrak{F}$ be a filter on X. A point $y \in Y$ is said to be a limit point (or simply a limit) (resp. cluster point) of f with respect to the filter $\mathfrak{F}$ if y is a limit point (resp. cluster point) of the filter base $f(\mathfrak{F})$ .

The relation "y is a limit of f with respect to the filter $\mathfrak{F}$ " is written $\lim_{\mathfrak{F}} f = y$ , or $\lim_{x, \mathfrak{F}} f(x) = y$ , or $\lim_x f(x) = y$ if there is no risk of confusion.

From Definition 3 and Propositions 1 (no. 1) and 3 (no. 2) we deduce the following criteria:

Proposition 7

A point $y \in Y$ is a limit of $f$ with respect to the filter $\mathfrak{F}$ if and only if, for each neighbourhood $V$ of $y$ in $Y$ , there is a set $M \in \mathfrak{F}$ such that $f(M) \subset V$ (i.e. $\overline{f}(V) \in \mathfrak{F}$ for each neighbourhood $V$ of $y$ ).* *A point $y \in Y$ is a cluster point of $f$ with respect to $\mathfrak{F}$ if and only if for each neighbourhood $V$ of $y$ and each $M \in \mathfrak{F}$ there is a point $x \in M$ such that $f(x) \in V$ .

Example 1

A sequence of points $(x_n)_{n \in \mathbf{N}}$ of a topological space is a mapping $n \to x_n$ of $\mathbf{N}$ into $X$ . In analysis one frequently uses the notions of limit point and cluster point of such a mapping with respect to the Fréchet filter (§ 6, no. 1) on $\mathbf{N}$ ; if $y$ is a limit of $n \to x_n$ with respect to the Fréchet filter, $y$ is said to be a limit of the sequence $(x_n)$ as n tends to infinity, and we write $\lim_{n \to \infty} x_n = y$ . A cluster point of the mapping $n \to x_n$ with respect to the Fréchet filter is called a cluster point of the sequence $(x_n)$ .

Thus a point $y \in X$ is a limit (resp. cluster point) of a sequence $(x_n)$ of points of $X$ if it is a limit point (resp. cluster point) of the elementary filter associated with $(x_n)$ (§ 6, no. 8).

The point $y$ is a limit of a sequence $(x_n)$ in $X$ if and only if, for every neighbourhood $V$ of $y$ in $X$ , all but a finite number of the terms of the sequence $(x_n)$ are in $V$ , i.e. there is an integer $n_0$ such that $x_n \in V$ for all $n \geq n_0$ . Likewise $y$ is a cluster point of the sequence $(x_n)$ if and only if, for every neighbourhood $V$ of $y$ in $X$ and every integer $n_0$ , there is an integer $n \geq n_0$ such that $x_n \in V$ .

Example 2

More generally, let $f$ be a mapping of a directed set $A$ into a topological space $X$ . If $x \in X$ is a limit (resp. cluster point) of $f$ with respect to the section filter of $A$ then $x$ is said to be a limit (resp. cluster point) of $f$ with respect to the directed set $A$ , and we write $x = \lim_{z \in A} f(z)$ .

If $y$ is a limit (resp. cluster point) of a mapping $f : X \to Y$ with respect to a filter $\mathfrak{F}$ on $X$ , then $y$ remains a limit (resp. cluster point) of $f$ with respect to $\mathfrak{F}$ if we replace the topology of $Y$ by a coarser topology, or if we replace the filter $\mathfrak{F}$ by a finer (resp. coarser) filter.

Proposition 8

Let $f$ be a mapping of a set $X$ into a topological space $Y$ ; then $y \in Y$ is a cluster point of $f$ with respect to $\mathfrak{F}$ if and only if there is a filter $\mathfrak{G}$ on $X$ which is finer than $\mathfrak{F}$ and such that $y$ is a limit of $f$ with respect to $\mathfrak{G}$ .

For if $y$ is a cluster point of $f$ with respect to $\mathfrak{F}$ , and if $\mathfrak{B}$ is the neighbourhood filter of $y$ , then $\overline{f}^{-1}(\mathfrak{B})$ is a filter base on $X$ since every set of $\overline{f}^{-1}(\mathfrak{B})$ meets every set of $\mathfrak{F}$ (§ 6, no. 6). This remark shows also that there is a filter $\mathfrak{G}$ on $X$ which is finer than both $\mathfrak{F}$ and the filter with base $\overline{f}^{-1}(\mathfrak{B})$ (§ 6, no. 2, Proposition 1, Corollary 2), hence that $y$ is a limit point of $f$ with respect to $\mathfrak{G}$ .

Notice finally that if $f$ is a mapping of a set $X$ into a topological space $Y$ , the set of cluster points of $f$ with respect to a filter $\mathfrak{F}$ on $X$ is closed in $Y$ (no. 2, Proposition 5) and possibly empty.

Remark

If $y \in Y$ is a limit (resp. cluster point) of a mapping $f : X \to Y$ with respect to a filter $\mathfrak{F}$ on $X$ , then $y$ is also a limit (resp. cluster point) of every function $g : X \to Y$ which has the same germ as $f$ with respect to $\mathfrak{F}$ (§ 6, no. 9); $y$ is said to be a limit (resp. cluster point) of the germ $\tilde{f}$ of $f$ with respect to $\mathfrak{F}$ .

4. Limits and continuity

Let $X, Y$ be two topological spaces, $f$ a mapping of $X$ into $Y$ , $\mathfrak{B}$ the neighbourhood filter in $X$ of a point $a \in X$ . Instead of saying that $y \in Y$ is a limit of $f$ with respect to the filter $\mathfrak{B}$ and writing $y = \lim_{\mathfrak{B}} f$ , we use the special notation

y = \lim_{i \in I} f(x),

and we say that $y$ is a limit of $f$ at the point $a$ , or that $f(x)$ tends to $y$ as $x$ tends to $a$ . Similarly, instead of saying that $y$ is a cluster point of $f$ with respect to $\mathfrak{B}$ , we say that $y$ is a cluster point of $f$ at the point $a$ .

A consideration of the definition of continuity (§ 2, no. 1, Definition 1) Proposition 7 of no. 3 shows that:

Proposition 9

A mapping* $f$ of a topological space $X$ into a topological space $Y$ is continuous at a point $a \in X$ if and only if $\lim_{i \in I} f(x) = f(a)$ .

Corollary 1

Let* $X, Y$ be two topological spaces, $f$ a mapping of $X$ into $Y$ which is continuous at a point $a \in X$ ; then, for every filter base $\mathfrak{B}$ on $X$ which converges to $a$ , the filter base $f(\mathfrak{B})$ converges to $f(a)$ . Conversely if, for every ultrafilter $\mathfrak{U}$ on $X$ which converges to $a$ , the ultrafilter base $f(\mathfrak{U})$ converges to $f(a)$ , then $f$ is continuous at $a$ .

The first assertion is an immediate consequence of Proposition 9. To prove the second, suppose that $f$ is not continuous at $a$ ; then there is a neighbourhood W of $f(a)$ in Y such that $\overline{f}^{-1}(W)$ does not belong to the filter $\mathfrak{B}$ of neighbourhoods of $a$ in X. Hence (§ 6, no. 4, Proposition 7) there is an ultrafilter $\mathfrak{U}$ , finer than $\mathfrak{B}$ , which does not contain $\overline{f}^{-1}(W)$ and therefore contains its complement $A = X - \overline{f}^{-1}(W)$ (§ 6, no. 4, Proposition 5); since $f(A) \cap W = \varnothing$ , $f(\mathfrak{U})$ does not converge to $f(a)$ .

Corollary 2

Let g be a mapping of a set Z into a topological space X, which has a limit a with respect to a filter $\mathfrak{F}$ on Z; then if the map $f : X \to Y$ is continuous at a, the composition $f \circ g$ has $f(a)$ as a limit point with respect to $\mathfrak{F}$ .

5. Limits relative to a subspace

Let X, Y be two topological spaces, let A be a subset of X, and let $a \in X$ be a point of the closure of A (but not necessarily in A). Let $\mathfrak{F}$ be the trace on A of the neighbourhood filter of $a$ in X. If $f$ is a mapping of A into Y, then instead of saying that $y \in Y$ is a limit of $f$ with respect to $\mathfrak{F}$ and writing $y = \lim_{\mathfrak{F}} f$ , we write

y = \lim_{x \to a, x \in A} f(x)

and we say that $y$ is a limit of f at a, relative to the subspace A, or that $f(x)$ tends to y as x tends to a while remaining in A. We have then $y \in f(A)$ .

If $A = \mathbf{C}\{a\}$ where $a$ is not an isolated point of X, then we write

y = \lim_{x \to a, x \neq a} f(x)

instead of $y = \lim_{x \to a, x \in A} f(x)$ .

We make analogous definitions for cluster points.

If $f$ is the restriction to A of a mapping $g : X \to Y$ , we say that $g$ has a limit (resp. cluster point) $y$ relative to A at a point $a \in \overline{A}$ , if $y$ is a limit (resp. cluster point) of $f$ at $a$ , relative to A.

Let B be a subset of A and let $a \in X$ be a point of the closure of B; if $y$ is a limit at $a$ , relative to A, of a map $f : A \to Y$ , then $y$ is also a limit of $f$ at $a$ , relative to B; the converse is not necessarily true.

But if V is a neighbourhood in X of a point $a \in \overline{A}$ and if $f$ has a limit $y$ at $a$ , relative to $V \cap A$ , then $y$ is still a limit of $f$ at $a$ , relative to A.

Let $a$ be a non-isolated point of X, so that $a$ is in the closure of $\mathbf{C}\{a\}$ . Then a mapping $f : X \to Y$ is continuous at $a$ if and only if we have $f(a) = \lim_{x \to a, x \neq a} f(x)$ ; this follows immediately from the definitions.

6. Limits in product spaces and quotient spaces

Proposition 10

Let $X$ be a set, let $(Y_i)_{i \in I}$ be a family of topological spaces, and for each $i \in I$ let $f_i$ be a mapping of $X$ into $Y_i$ . Let $X$ be given the coarsest topology $\mathfrak{T}$ for which the $f_i$ are continuous. Then a necessary and sufficient condition for a filter $\mathfrak{F}$ on $X$ to converge to $a \in X$ is that for each $i \in I$ the filter base $f_i(\mathfrak{F})$ should converge to $f_i(a)$ in $Y_i$ .

The condition is necessary since the $f_i$ are continuous (no. 4, Proposition 9, Corollary 1). Conversely, suppose that the condition is satisfied, and let $V$ be an open neighbourhood of $a$ in $X$ . By the definition of $\mathfrak{T}$ (§ 2, no. 3, Proposition 4) there is a finite subset $J$ of $I$ , and for each $i \in J$ an open subset $U_i$ of $Y_i$ , such that $f_i(a) \in U_i$ for $i \in J$ and such that $V$ contains the set

\bigcap_{i \in J} f_i^{-1}(U_i).

The hypothesis implies that

f_i^{-1}(U_i) \in \mathfrak{F}

(no. 3, Proposition 7); since

J

is finite, it follows that

M = \bigcap_{i \in J} f_i^{-1}(U_i)

belongs to

\mathfrak{F}

, and

M \subset V

. This completes the proof.

Corollary 1

A filter $\mathfrak{F}$ on a product space $X = \prod_{i \in I} X_i$ converges to a point $x$ if and only if for each $i \in I$ the filter base $\operatorname{pr}_i(\mathfrak{F})$ converges to $\operatorname{pr}_i(x)$ .

Corollary 2

Let $f = (f_i)$ be a mapping of a set $X$ into a product space $Y = \prod_{i \in I} Y_i$ . Then $f$ has a limit $y = (y_i)$ with respect to a filter $\mathfrak{F}$ on $X$ if and only if for each $i \in I$ $f_i$ has limit $y_i$ with respect to $\mathfrak{F}$ .

Proposition 11

Let $R$ be an open equivalence relation on a topological space $X$ and let $\varphi$ be the canonical mapping $X \to X/R$ . Then for each $x \in X$ and each filter base $\mathfrak{B}'$ on $X/R$ which converges to $\varphi(x)$ , there is a filter base $\mathfrak{B}$ on $X$ which converges to $x$ and is such that $\varphi(\mathfrak{B})$ is equivalent to $\mathfrak{B}'$ .

If $U$ is any neighbourhood of $x$ in $X$ , then $\varphi(U)$ is a neighbourhood of $\varphi(x)$ in $X/R$ (§ 5, no. 3, Proposition 5), hence there is a set $M' \in \mathfrak{B}'$ such that $M' \subset \varphi(U)$ ; if we put $M = U \cap \varphi^{-1}(M')$ , then $M' = \varphi(M)$ . This shows that as $M'$ runs through $\mathfrak{B}'$ and $U$ runs through the neighbourhood filter of $x$ , the sets $U \cap \varphi^{-1}(M')$ form a filter base $\mathfrak{B}$ on $X$ ; clearly $\mathfrak{B}$ converges to $x$ and $\varphi(\mathfrak{B})$ is equivalent to $\mathfrak{B}'$ .

Exercises

See the exercises for § 7.