Limits and Continuous Functions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 25 January 2014

Filters and limits

Let $X$ be a set.

• A filter on $X$ is a collection $ℱ$ of subsets of $X$ such that

  (a)   (upper ideal) if $N\in ℱ$ and $N\subseteq E\subseteq X$ then $E\in ℱ$,

  (b)   (closed under finite intersections) if $N_1,N_2,\dots ,N_l\in ℱ$ then $N_1\cap N_2\cap \cdots \cap N_l\in ℱ$, and

  (c)   $\varnothing \notin ℱ$.

• An ultrafilter on $X$ is a maximal filter $ℱ$ on $X$ (with respect to inclusion).

Let $Y$ be a topological space. Let $y\in Y$.

• A neighbourhood of $y$ is a subset $N\subseteq Y$ such that there exists an open set $U$ with $y\in U\subseteq N$.
• The neighbourhood filter of $y$ is

  $𝒩\left(y\right)=\left\{\text{neighborhoods of}y\right\}$.

Let $Y$ be a topological space and let $𝒢$ be a filter on $Y$.

• A limit point of $𝒢$ is a point $y\in Y$ such that $𝒢\supseteq 𝒩\left(y\right).$
• A cluster point of $𝒢$ is a point $y\in Y$ such that $y\in \left(\bigcap _{C\in 𝒢}\bar{C}\right),$ where $\bar{C}$ is the closure of $C$.

The issue of which properites of a space guarantee that limit points are unique (when they exist) is treated in Section ??: Hausdorff spaces, and the the issue of which properties of a space guarantee that cluster points exist is treated in Section ??: Compact spaces.

HW: Let $Y$ be a topological space and let $𝒢$ and $ℱ$ be filters on $Y$ such that $𝒢\supseteq ℱ$.

1. (a) Show that if $y$ is a limit point of $ℱ$ then $y$ is a limit point of $𝒢$.
2. (b) Show that if $y$ is a cluster point of $𝒢$ then $y$ is a cluster point of $ℱ$.

HW: Let $Y$ be a topological space and let $𝒢$ be a filter on $Y$. Let $y\subseteq Y$. Show that $y$ is a cluster point of $𝒢$ if and only if there exists a filter $ℱ$ on $Y$ such that $ℱ\supseteq 𝒢$ and $ℱ\supseteq 𝒩\left(y\right)$.

HW: Let $Y$ be a topological space and let $𝒢$ be a filter on $Y$. Let $y\subseteq Y$. Show that $y$ is a cluster point of $𝒢$ if and only if there exists a filter $ℱ\supseteq 𝒢$ such that $y$ is a limit point of $ℱ$.

HW: Let $Y$ be a topological space and let $𝒢$ be a filter on $Y$. Let $y\subseteq Y$. Show that if $y$ is a limit point of $𝒢$ then $y$ is a cluster point of $𝒢$.

HW: Let $Y$ be a topological space and let $𝒢$ be an ultrafilter on $Y$. Let $y\subseteq Y$. Show that $y$ is a limit point of $𝒢$ if and only if $y$ is a cluster point of $𝒢$.

Let $X$ be a set with a filter $ℱ$ and let $Y$ be a topological space. Let $f:X\to Y$ be a function.

• A limit point of $f$ is a limit point of the coarsest filter containing $f\left(ℱ\right)$.
• A cluster point of $f$ is a cluster point of the coarsest filter containing $f\left(ℱ\right)$.

  Write    $y=\underset{ℱ}{\lim }f\left(x\right)$,    if $y$ is a limit point of $f$.

Sequences

Let $Y$ be a topological space.

• A sequence $(y_1,y_2,y_3,\dots )$ in $Y$ is a function $$\begin{array}{rcl}{\mathbb{Z}}_{>0}& ⟶& Y\\ n& ⟼& y_n.\end{array}$$
• The filter associated to $(y_1,y_2,\dots )$ is the filter $$ℱ(y_1,y_2,\dots )=\{E\subseteq Y|E\text{contains all but a finite number of}{y}_1,y_2,\dots \}$$
• A limit of the sequence $(y_1,y_2,y_3,\dots )$ is a limit point of the sequence with respect to $ℱ(y_1,y_2,\dots )$.
• A cluster point of the sequence $(y_1,y_2,y_3,\dots )$ is a cluster point of the sequence with respect to $ℱ(y_1,y_2,\dots )$.
• The Fréchet filter on ${\mathbb{Z}}_{>0}$ is $ℱ=\left\{\text{complements of finite subsets of}{\mathbb{Z}}_{>0}\right\}$.
• A limit of the sequence $(y_1,y_2,y_3,\dots )$ is a limit point of the sequence with respect to the Fréchet filter on ${\mathbb{Z}}_{>0}$.
• A cluster point of the sequence $(y_1,y_2,y_3,\dots )$ is a cluster point with respect to the Fréchet filter on ${\mathbb{Z}}_{>0}$.

Write     $y=\underset{n\to \infty }{\lim }{y}_n$,    if $y$ is a limit point of the sequence $\left(y_1,y_2,y_3,\dots \right)$.

Let $Y$ be a topological space and let $\left(y_1,y_2,y_3,\dots \right)$ be a sequence in $Y$. Let $y\in Y$.

1. $y$ is a limit point of $\left(y_1,y_2,y_3,\dots \right)$ if and only if
   1. if $N_y$ is a neighbourhood of $y$ then there exists $n_0\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>n_0$ then $y_n\in N_y$.
2. $y$ is a cluster point of $\left(y_1,y_2,y_3,\dots \right)$ if and only if
   1. if $N_y$ is a neighbourhood of $y$ and $n_0\in {\mathbb{Z}}_{>0}$ then there exists $n\in {\mathbb{Z}}_{>0}$ such that $n>n_0$ and $y_n\in N_y$.

		Proof.
	⇒) Assume $y$ is a limit point of $(y_1,y_2,...).$ So $$y=\underset{n\to \infty }{\lim }{y}_n=\underset{ℱ}{\lim }f\left(n\right),$$ where $$\begin{array}{rrcl}f:& {\mathbb{Z}}_{>0}& \to & y\\ & n& \mapsto & y_n\end{array}$$ and $ℱ$ is the Fréchet filter on ${\mathbb{Z}}_{>0}.$ So $𝒩\left(y\right)\subseteq ⟨f\left(ℱ\right)⟩.$ So $𝒩\left(y\right)\subseteq ⟨f\left(\genfrac{}{}{0ex}{}{\text{complements of}}{\text{finite subsets of}{\mathbb{Z}}_{>0}}\right)⟩.$ To show: If $N_y$ is a neighbourhood of $y$ then there exists $n_0\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>n_0$ then $y_n\in N_y.$ Assume $N_y$ is a neighbourhood of $y.$ So $N_y\in 𝒩\left(y\right).$ So there exists a finite set $S\subseteq {\mathbb{Z}}_{>0}$ such that $N_y\supseteq f\left(S^c\right).$ Let $n_0=\sup \left(S\right).$ To show: If $n\in {\mathbb{Z}}_{>0}$ and $n>n_0$ then $y_n\in N_y.$ Assume $n\in {\mathbb{Z}}_{>0}$ and $n>n_0.$ Then $n\in S^c.$ So $y_n=f\left(n\right)\in f\left(S^c\right).$ Since $f\left(S^c\right)\subseteq N_y, y_n\in N_y.$ ⇐) Assume that if $N_y$ is a neighbourhood of $y$ then there exists $n_0\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>n_0$ then $y_n\in N_y.$ To show: $$y=\underset{n\to \infty }{\lim }{y}_n.$$ To show: $$𝒩\left(y\right)\subseteq ⟨f\left(\genfrac{}{}{0ex}{}{\text{complements of}}{\text{finite subsets of}{\mathbb{Z}}_{>0}}\right)⟩.$$ To show: If $N_y\in 𝒩\left(y\right)$ then there exists a finite subset $S\subseteq {\mathbb{Z}}_{>0}$ such that $N\supseteq f\left(S^c\right).$ Assume $N_y\in 𝒩\left(y\right).$ To show: There exists a finite subset $S\subseteq {\mathbb{Z}}_{>0}$ such that $N_y\supseteq f\left(S^c\right).$ Let $n_0\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>n_0$ then $y_n\in N_y.$ Let $S=\{1,2,...,n_0\}$ so that $S$ is a finite subset of ${\mathbb{Z}}_{>0}.$ Then $$\begin{array}{rcl}f\left(S^c\right)& =& f\left(\{n_0+1,n_0+2,...\}\right)\\ & =& \{y_{n_0+1},y_{n_0+2},...\}\\ & =& \left\{y_n \right| n>n_0\}\subseteq N_y.\end{array}$$ ⇒) Assume $y$ is a cluster point of $(y_1,y_2,...).$ To show: If $N_y$ is a neighbourhood of $y$ and $n_0\in {\mathbb{Z}}_{>0}$ then there exists $n\in {\mathbb{Z}}_{>0}$ such that $n>n_0$ and $y_n\in N_y.$ We know: If $U\in ⟨f\left(\genfrac{}{}{0ex}{}{\text{complements of}}{\text{finite subsets of}{\mathbb{Z}}_{>0}}\right)⟩.$ then $y\in \stackrel{\_}{U}.$ Assume $N_y$ is a neighbourhood of $y$ and $n_0\in {\mathbb{Z}}_{>0}.$ To show: There exists $n\in {\mathbb{Z}}_{>0}$ such that $n>n_0$ and $y_n\in N_y.$ To show: There exists $n\in {\{1,...,n_0\}}^c$ such that $y_n\in N_y.$ To show: $$f\left({\{1,...,n_0\}}^c\right)\cap N_y\ne \varnothing .$$ We know: $$y\in \stackrel{\_}{f\left({\{1,...,n_0\}}^c\right)}.$$ So $$N_y\cap f\left({\{1,...,n_0\}}^c\right)\ne \varnothing .$$ ⇐) Assume that if $N_y$ is a neighbourhood of $y$ and $n_0\in {\mathbb{Z}}_{>0}$ then there exists $n\in {\mathbb{Z}}_{>0}$ such that $n>n_0$ and $y_n\in N_y.$ To show: If $U\in ⟨f\left(\genfrac{}{}{0ex}{}{\text{complements of}}{\text{finite subsets of}{\mathbb{Z}}_{>0}}\right)⟩.$ then $y\in \stackrel{\_}{U}.$ Assume $U\in ⟨f\left(\genfrac{}{}{0ex}{}{\text{complements of}}{\text{finite subsets of}{\mathbb{Z}}_{>0}}\right)⟩.$ So $U\supseteq f\left(S^c\right)$ for a finite set $S\subseteq {\mathbb{Z}}_{>0}.$ Let $n_0=\sup \left(S\right).$ Then $U\supseteq f\left(S^c\right)\supseteq \{y_{n_0+1},y_{n_0+2},...\}=\left\{y_n \right| n>n_0\}.$ To show: $y\in \stackrel{\_}{U}.$ To show: If $N_y$ is a neighbourhood of $y$ then $N_y\cap U\ne \varnothing .$ Assume $N_y$ is a neighbourhood of $y.$ Since $N_y\cap \left\{y_n \right| n>n_0\}\ne \varnothing$ and $\left\{y_n \right| n>n_0\}\subseteq U,$ then $N_y\cap U\ne \varnothing .$ $\square$

Example: The sequence $$y_n=\{\begin{array}{ll}1-\frac{1}{n},& \text{if}n\text{is even,}\\ 0,& \text{if}n\text{is odd,}\end{array}\text{in}ℝ$$ has no limit point, but has cluster points at 1 and 0.

Limits and continuity

Let $X$ and $Y$ be topological spaces and let $f:X\to Y$ be a function. Let $a\in X$.
A limit of $f$ as $x$ approaches $a$ is a limit point of $f$ with respect to the neighbourhood filter $𝒩\left(a\right)$.

Write     $y=\underset{x\to a}{\lim }f\left(x\right)$,    if $y$ is a limit of $f$ as $x$ approaches $a$.

The function $f:X\to Y$ is continuous at $x=a$ if it satisfies

1. if $N$ is a neighbourhood of $f\left(a\right)$ in $Y$ then $f^{-1}\left(N\right)$ is a neighbourhood of $a$ in $X$.

Let $X$ and $Y$ be topological spaces and let $a\in X$. A function $f:X\to Y$ is continuous at $x=a$ if and only if $$\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right).$$

		Proof.
	⇒) Assume $f:X\to Y$ is continuous at $x=a.$ To show: $$\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$$ To show: $$𝒩\left(f\left(a\right)\right)\subseteq ⟨f\left(𝒩\left(a\right)\right)⟩.$$ To show: If $N\in 𝒩\left(f\left(a\right)\right)$ then there exists $U\in 𝒩\left(a\right)$ such that $f\left(U\right)\subseteq N.$ Assume $N\in 𝒩\left(f\left(a\right)\right).$ To show: There exists such that $f\left(U\right)\subseteq N.$ Let $U=f^{-1}\left(N\right).$ Since $f$ is continuous at $x=a,$ $U\in 𝒩\left(a\right).$ To show: $f\left(U\right)\subseteq N.$ Since $f\left(U\right)=f\left(f^{-1}\left(N\right)\right)=N, f\left(U\right)\subseteq N.$ So $$\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right).$$ ⇐) Assume $$f\left(a\right)=\underset{x\to a}{\lim }f\left(x\right).$$ So $𝒩\left(f\left(a\right)\right)\subseteq ⟨f\left(𝒩\left(a\right)\right)⟩.$ To show: If $N\in 𝒩\left(f\left(a\right)\right)$ then $f^{-1}\left(N\right)\in 𝒩\left(a\right).$ Assume $N\in 𝒩\left(f\left(a\right)\right).$ To show: $f^{-1}\left(N\right)\in 𝒩\left(a\right).$ Since $𝒩\left(f\left(a\right)\right)\subseteq ⟨f\left(𝒩\left(a\right)\right)⟩$ we know: There exists $U\in 𝒩\left(a\right)$ such that $N\supseteq f\left(U\right).$ So $f^{-1}\left(N\right)\supseteq U.$ So $f^{-1}\left(N\right)\in 𝒩\left(a\right).$ $\square$

Generating filters

Let $X$ be a set and let ${ℱ}_1$ and ${ℱ}_2$ be filters on $X$. The filter ${ℱ}_1$ is finer than ${ℱ}_2$ if ${ℱ}_1\supseteq {ℱ}_2$.

If $ℬ$ is a collection of subsets of $X$ that satisfies

(a)   if $B_1,B_2\in ℬ$ then there exists $B\in ℬ$ such that $B\subseteq B_1\cap B_2$, and

(b)   $ℬ\ne \varnothing$ and $\varnothing \notin ℬ$,

then

$ℱ=\left\{\text{<em>subsets of</em>}X\text{<em>that contain a set in</em>}ℬ\right\}$

is the coarsest filter containing $ℬ$.

Let $X$ be a set and let $𝒮$ be a collection of subsets of $X$. If

$ℬ=\left\{\text{finite intersections of elements of}𝒮\right\}$     and    $ℱ=\left\{\text{subsets of}X\text{containing a set in}ℬ\right\}$

then $ℬ$ satisfies (a) and (b) and $ℱ$ is the coarsest filter on $X$ containing $𝒮$.

Metric spaces

Let $X$ and $Y$ be metric spaces and let $f:X\to Y$ be a function. Let $a\in X$ and $y\in Y$. Then $$\underset{x\to a}{\lim }f\left(x\right)=y\text{if and only if}$$ if $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that $$\text{if}d(x,a)<\delta \text{then}d(f\left(x\right),y)<\epsilon .$$

Let $Y$ be a metric space with the metric space uniformity.
Let $(y_1,y_2,\dots )$ be a sequence in $Y$ and let $ℱ(y_1,y_2,\dots )$ be the filter associated to $(y_1,y_2,\dots )$. Then $$(y_1,y_2,\dots )\text{is a Cauchy sequence in}Y\text{if and only if}ℱ(y_1,y_2,\dots )\text{is a Cauchy filter on}Y,$$

Notes and References

Historically, the mathematical community became infatuated by limits, partly because of the many applications of "calculus" and the ideas of infinitesimals, but also partly because they weren't very well understood. This infatuation has often focused on the epsilon-delta definition of limits, which has advantages and disadvantages.

The thorough and easy definition of limits given in this exposition, not depending on a metric space, follows [Bou, Top., Ch.I, §6-7]. In particular, the definition of filter, neighbourhood filter, and Fréchet filter, are in [Bou, Top., Ch.I, §6 No 1] and the definition of limit point and cluster point of a filter are [Bou, Top., Ch.I, §7 Def. 1,2] and the definition of limit point and cluster point of a function are [Bou, Top., Ch.I, §7 Def 3].

Theorem 2.1 is [Bou, Top., Ch.I, §7 Prop 9] and Theorem 3.1 is Example 1 in [Bou, Top., Ch.I, §7 No 3]. The discussion in §4 Generating filters is based on [Bou, Top., Ch.I, §6 No. 2-3].

The second and third conditions in the definition of a filter say that finite intersections of elements of a filter cannot be empty. This is the rigidity condition that plays an important role in arguments about limit points and compactness.

A point $y$ is a cluster point of $𝒢$ if $y$ is a close point for all the sets $C\in 𝒢$.

The partial order on filters by inclusion is fundamental. A point $y$ is a cluster point if the supremum of $𝒢$ and the neighborhood filter of $y$ exists. If filters are analogous to sequences then inclusion of filters is analogous to subsequences. In particular, the existence of an ultrafilter with a limit point is analogous to a subsequence with a limit point.

References

[Bou] N. Bourbaki, General Topology, Springer-Verlag, 1989. MR1726779.

[BR] W. Rudin, Principles of mathematical analysis, Third edition, International Series in Pure and Applied Mathematics, McGraw-Hill 1976. MR0385023.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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