Preliminaries to Topology

Recall Def 143.1.

Definition 372.1(Definition).

If $X$ is any set, the collection of all subsets of $X$ is a topology on $X$, called the discrete topology. The collection consisting of $X$ and $\varnothing$ only is also a topology on $X$, called the trivial topology.

Definition 372.2(Definition).

Suppose $\mathcal{T}$ and ${\mathcal{T}}^{\prime }$ are two topologies on a given set $X$. If ${\mathcal{T}}^{\prime }\supseteq \mathcal{T}$, we say that ${\mathcal{T}}^{\prime }$ is finer than $\mathcal{T}$. If ${\mathcal{T}}^{\prime }\subseteq \mathcal{T}$, we say that ${\mathcal{T}}^{\prime }$ is coarser than $\mathcal{T}$.

Definition 372.3(Basis).

If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ (called basis elements) such that

1. For each $x\in X$, there is at least one $B\in \mathcal{B}$ such that $x\in B$.
2. If $x\in B_1\cap B_2$, then there exists $B_3\in \mathcal{B}$ such that $x\in B_3$ and $B_3\subseteq B_1\cap B_2$.

If $\mathcal{B}\subseteq 2^X$ satisfies these conditions, we define the topology generated by $\mathcal{B}$ to be all sets obtained by arbitrary unions of its elements ¹. Given a topology $\mathcal{T}$ for $X$, $\mathcal{B}$ is said to be a basis for $\mathcal{T}$ if the topology generated by $\mathcal{B}$ is equal to $\mathcal{T}$.

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If the second requirement is relaxed in Def 3, we obtain what is called a subbasis.

Definition 372.4(Subbasis).

A subbasis for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by a subbasis $S$ is defined to be the topology generated by the basis whose elements are all finite intersections of elements of $S$.

==The topology generated by a subbasis $S$ is the coarsest topology on $X$ in which all the elements of $S$ are open. ==

The following lemma lets us obtain a basis for a given topology.

Lemma 372.5(Munkres (2000) 13.2).

Let $(X,\mathcal{T})$ be a topological space. Suppose $\mathcal{C}\subseteq \mathcal{T}$ is such that of each $U\in \mathcal{T}$ and $x\in U$, there is an element $C\in \mathcal{C}$ such that $x\in C\subseteq U$. Then $\mathcal{C}$ is a basis for the topology $X$.

Proof.

That $\mathcal{C}$ satisfies Def 3 is clear. Let ${\mathcal{T}}^{\prime }$ be the topology generated by $\mathcal{C}$. Since $\mathcal{C}\subseteq \mathcal{T}$, it follows that ${\mathcal{T}}^{\prime }\subseteq \mathcal{T}$. It is also clear that any $U\in \mathcal{T}$ can be expressed as a union of elements of $\mathcal{C}$, so $\mathcal{T}\subseteq {\mathcal{T}}^{\prime }$.□

When topologies are given by bases, the following lemma lets us determine weather one topology is finer than the other.

Lemma 372.6(Munkres (2000) 13.3).

Let $\mathcal{B}$ and ${\mathcal{B}}^{\prime }$ be bases for the topologies $\mathcal{T}$ and ${\mathcal{T}}^{\prime }$, respectively, on $X$. TFAE:

1. ${\mathcal{T}}^{\prime }$ is finer than $\mathcal{T}$.

2. For each $x\in X$ and each $B\in \mathcal{B}$ containing $x$, there exists $B^{\prime }\in {\mathcal{B}}^{\prime }$ such that $x\in B^{\prime }\subseteq B$.

Proof.

$(1⟹2)$ Take $B^{\prime }=B$.
$(2⟹1)$ Let $U\in \mathcal{T}$. By $(2)$, $U$ can clearly be expressed as a union of elements of ${\mathcal{T}}^{\prime }$.□

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The order topology

Definition 372.7(Order topology).

Let $X$ be a set with a simple order relation; assume $X$ has more than one element. Let $\mathcal{B}$ be the collection of all sets of the following types:

1. All “open intervals” $(a,b)=\{x:a<x<b\}$ in $X$.
2. All intervals of the form $[a_0,b)=\{x:a_0⩽x<b\}$, where $a_0$ is the smallest element (if any) of $X$.
3. All intervals of the form $(a,b_0]=\{x:a<x⩽b_0\}$, where $b_0$ is the largest element (if any) of $X$.

The collection $\mathcal{B}$ is a basis for a topology on $X$, called the order topology.

Open rays of $X$ form a subbasis for the order topology on $X$.

The Subspace topology

Definition 372.8(Subspace topology).

Let $(X,\mathcal{T})$ be a topological space. If $Y\subseteq X$, the collection

$${\mathcal{T}}_Y:=\{Y\cap U:U\in \mathcal{T}\}$$

is a topology on $Y$, called the subspace topology. With this topology, $Y$ is called a subspace of $X$.

If $Y$ is a subspace of $X$, $A\subseteq Y$ is open/closed in $Y$ $⟺$ $A=S\cap Y$ for $S$ open/closed in $X$.

Lemma 372.9(Munkres (2000) 16.1, 16.2).

1. If $\mathcal{B}$ is a basis for the topology of $X$ then the collection ${\mathcal{B}}_Y:=\{B\cap Y:B\in \mathcal{B}\}$ is a basis for the subspace topology on $Y$.
2. Let $Y$ be a subspace of $X$. If $U$ is open in $Y$ and $Y$ is open in $X$, then $U$ is open in $X$.

Proposition 372.10(Munkres (2000) 16.3).

If $A$ is a subspace of $X$ and $B$ is a subspace of $Y$, then the product topology on $A\times B$ is the same as the topology $A\times B$ inherits as a subspace of $X\times Y$.

Limit points and convergence

Definition 372.11(Definition).

Given $A\subseteq X$, the interior of $A$ is defined as the union of all open sets contained in $A$, and the closure of $A$ is defined as the intersection of all closed sets containing $A$.

Proposition 372.12(Proposition).

Let $A\subseteq Y\subseteq X$. Then, (closure of $A$ in $Y$)=(closure of $A$ in $X$)$\cap Y$.

Proposition 372.13(Munkres (2000) 17.5).

Let $A\subseteq X$.

1. $x\in \overline{A}$ iff every open $U\ni x$ intersects $A$.
2. If $\mathcal{B}$ is a basis for the topology on $X$, then $x\in \overline{A}$ iff every basis element $B\in \mathcal{B}$ containing $x$ intersects $A$.

Definition 372.14(Limit points).

Let $A\subseteq X$, $x\in X$. We say $x$ is a limit point of $A$ if $x\in \overline{A\setminus \{x\}}$.

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Clearly, $\overline{A}=A\cup A^{\prime }$, where $A^{\prime }$ is the set of all limit points of $A$. Thus, $A\subseteq X$ is closed iff it contains all its limit points.

Definition 372.15(Convergence).

$\{x_n\}\subseteq X$ is said to converge to $x\in X$ if for each neighborhood $U$ of $x$, there is a positive integer $N$ such that $x_n\in N$ for all $n⩾N$.

Continuity

Definition 372.16(Continuity).

$f:X\to Y$ is said to be continuous if for each open $V\subseteq Y$, $f^{-1}(V)$ is open in $X$.

Clearly, it suffices to show that the preimage of every basis (or subbasis) element is open to prove continuity.

Proposition 372.17(Proposition).

1. $f$ is continuous $⟺$ $f(\overline{A})\subseteq \overline{f(A)}$ for all $A\subseteq X$.
2. The map $f:X\to Y$ is continuous if $X$ can be written as the union of open sets $U_{\alpha }$ such that $f|{}_{U_{\alpha }}$ is continuous for each $\alpha$.

Proposition 372.18(The pasting lemma).

Let $X=A\cup B$, where $A$ and $B$ are closed in $X$. Let $f:A\to Y$ and $g:B\to Y$ be continuous. If $f(x)=g(x)$ for every $x\in A\cap B$, then $f$ and $g$ combine to give a continuous function $h:X\to Y$.

The product topology

Definition 372.19(Product topology).

Let $\{X_{\alpha }:\alpha \in J\}$ be an indexed family of topological spaces. Let ${\mathcal{S}}_{\beta }$ denote the collection

$${\mathcal{S}}_{\beta }:=\{{\pi }_{\beta }^{-1}(U_{\beta }):U_{\beta }\text{ open in }{X}_{\beta }\},$$

and let $\mathcal{S}$ denote the union of these collections,

$$\mathcal{S}=\bigcup _{\beta \in J}{\mathcal{S}}_{\beta }.$$

The topology generated by the subbasis $\mathcal{S}$ is called the product topology on ${\prod }_{\beta \in J}{X}_{\beta }$.

Theorem 372.20(Theorem).

Suppose the topology on each space $X_{\alpha }$ is given by a basis ${\mathcal{B}}_{\alpha }$. The collection of all sets of the form

$$\prod _{\alpha \in J}{B}_{\alpha },$$

where $B_{\alpha }\in {\mathcal{B}}_{\alpha }$ for finitely many indices $\alpha$ and $B_{\alpha }=X_{\alpha }$ for all the remaining indices, will serve as a basis for the product topology on ${\prod }_{\alpha \in J}{X}_{\alpha }$.

Proposition 372.21(Proposition).

If each space $X_{\alpha }$ is Hausdorff, then ${\prod }_{\alpha }{X}_{\alpha }$ is Hausdorff in both the box and product topologies.

Proposition 372.22(Proposition).

Let $\{X_{\alpha }\}$ be an indexed family of spaces; let $A_{\alpha }\subseteq X_{\alpha }$ for each $\alpha$. If $\prod X_{\alpha }$ is given either the product or the box topology, then

$$\prod {\overline{A}}_{\alpha }=\overline{\prod A_{\alpha }}.$$

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Theorem 372.23(Theorem).

Let $f:A\to {\prod }_{\alpha }{X}_{\alpha }$ be given by $f_{\alpha }$ in each slot. Let $\prod X_{\alpha }$ have the product topology. Then, $f$ is continuous iff $f_{\alpha }$ is continuous for each $\alpha$.

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Thm 23 is the motivating requirement for the definition of the product topology. Proving $(⟹)$ requires us to show that each of the projection functions are continuous. It is easily seen that the “product topology” $\mathcal{T}$ is the coarsest topology that makes this happen. In other words, any topology ${\mathcal{T}}^{\prime }$ that hopes to satisfy Thm 23 must contain $\mathcal{T}$. In particular, the box topology, being finer than the product topology, also makes the projections continuous. However, the box topology does not satisfy $(⟸)$. In fact, while proving $(⟸)$, we realize that no topology finer than the product topology works. This is as it should be, since products can be defined using a universal property, and thus should be unique up to isomorphism.

The metric topology

Definition 372.24(Metrizable space).

If $X$ is a topological space, $X$ is said to be metrizable if there exists a metric $d$ on the set $X$ that induces the topology of $X$. A metric space is a metrizable space $X$ together with a specific metric $d$ that gives the topology of $X$.

Proposition 372.25(Proposition).

Let $X$ be a metric space with metric $d$. Define the standard bounded metric $\overline{d}:X\times X\to \mathbb{R}$ by

$$\overline{d}(x,y)=\min \{d(x,y),1\}.$$

$\overline{d}$ is a metric that induces the same topology as $d$.

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Proposition 372.26(Proposition).

Let $d$ and $d^{\prime }$ be two metrics on the set $X$. Let $\mathcal{T}$ and ${\mathcal{T}}^{\prime }$ be the topologies they induce, respectively. Then ${\mathcal{T}}^{\prime }$ is finer than $\mathcal{T}$ iff for each $x\in X$ and each $ϵ>0$, there exists a $\delta >0$ such that $B_{d^{\prime }}(x,\delta )\subseteq B_d(x,ϵ)$.

Definition 372.27(Uniform metric).

Given an index set $J$, and given points $\mathbf{x},\mathbf{y}\in {\mathbb{R}}^J$, define the uniform metric $\overline{\rho }$ on ${\mathbb{R}}^J$ by the equation

$$\overline{\rho }(\mathbf{x},\mathbf{y})=\sup \{\overline{d}(x_{\alpha },y_{\alpha }):\alpha \in J\},$$

where $\overline{d}$ is the standard bounded metric on $\mathbb{R}$. The topology induced by $\overline{\rho }$ is called the uniform topology.

Proposition 372.28(Proposition).

The uniform topology on ${\mathbb{R}}^J$ is finer than the product topology and coarser than the box topology. These topologies are all different if $J$ is infinite.

In the case when $J$ is infinite, when is ${\mathbb{R}}^J$ is metrizable in either the box or product topology? Only when $J$ is countable and ${\mathbb{R}}^J$ has the product topology.

Proposition 372.29(Proposition).

Let $\overline{d}$ be the standard bounded metric on $\mathbb{R}$. For $\mathbf{x},\mathbf{y}\in {\mathbb{R}}^{\omega }$, define

$$D(\mathbf{x},\mathbf{y})=\sup \left\{\frac{\overline{d}(x_i,y_i)}{i}\right\}.$$

Then $D$ is a metric that induces the product topology on ${\mathbb{R}}^{\omega }$.

In general, it is true that countable products of metrizable spaces are metrizable.

Sequences and continuity

It would be nice if we could say the following: If $x$ lies in the closure of subset $A$ of the space $X$, then there exists a sequence of points of $A$ converging to $x$. This is not true in general, but it is true for metrizable spaces.

Definition 372.30(Countable basis at a point).

A space $X$ is said to have a countable basis at the point $x$ if there is a countable collection $\{U_n{\}}_{n\in \mathbb{N}}$ of neighborhoods of $x$ such that any neighborhood $U$ of $x$ contains at least one of the sets $U_n$.

Note that if $X$ has a countable basis at each of its points, $X$ is first countable. A metrizable space always satisfies the first countability axiom.

Proposition 372.31(The sequence lemma).

Let $X$ be a topological space.

1. Let $A\subseteq X$. If there is a sequence of points in $A$ converging to $x$, then $x\in \overline{A}$; the converse holds if $X$ has a countable basis at $x$.

2. Let $f:X\to Y$. If $f$ is continuous, then for every convergent sequence $\{x_n\}\to x$ in $X$, the sequence $\{f(x_n)\}$ converges to $f(x)$. Then converse holds if $X$ has a countable basis at $x$.

Proof.

$(1)$ The forward implication is evident. For the converse, suppose $x\in \overline{A}$ and $X$ has a countable basis at $x$. Fix an enumeration $\{U_n{\}}_{n=1}^{\infty }$ for the basis. This gives us a sequence of nested open sets: $U_1,U_1\cap U_2,U_1\cap U_2\cap U_3,\dots$ (these are nonempty since each $U_i$ contains $x$). Since $x\in \overline{A}$, we can choose $x_i\in U_1\cap \cdots \cap U_i$ for each $i\in \mathbb{N}$. Now suppose $V$ is any neighborhood of $x$, with $V\supseteq U_k$. Since $U_k\supseteq U_1\cap \cdots \cap U_k\supseteq U_1\cap \cdots \cap U_{k+1}\supseteq \dots$, it follows that $V$ contains the tail $x_k,x_{k+1},\dots$.

$(2)$ Again, the forward implication is easy. For the converse, show that $f(\overline{A})\subseteq \overline{f(A)}$.□

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To show that a space is not metrizable, we can show that the sequence lemma does not hold:

Example 372.32(Example).

${\mathbb{R}}^{\omega }$ in the box topology is not metrizable.

We show that the sequence lemma does not hold for $R^{\omega }$ using a argument reminiscent of diagonalization using the set $A=\{\mathbf{x}:x_i>0\forall i\in \mathbb{N}\}$ and the point $0=(0,0,\dots )$.

I’m pretty sure the same idea works to prove that $R^J$ for uncountable $J$ under the box topology is not metrizable; just work with a fixed countable subset of $J$.

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Example 372.33(Example).

An uncountable product of $\mathbb{R}$ with itself under the product topology is not metrizable.

Constructing continuous functions

Recall the uniform limit theorem.

Footnotes

1. That this prescription generates a topology should be obvious: $\mathcal{T}$ is clearly closed under arbitrary unions. For a finite intersection $S={\bigcap }_{i=1}^n\left({\bigcup }_{\alpha \in I_i}{U}_{\alpha }\right)$, $x\in S$ must lie in $U_{{\alpha }_1}\cap \cdots \cap U_{{\alpha }_n}\subseteq S$ for ${\alpha }_i\in I_i$. By definition, there exists a basis element $B_x\ni x$ contained in this finite intersection of basis elements. Let $S={\bigcup }_{x\in S}{B}_x$. ↩

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References

Munkres, J. R. (2000). Topology (2nd ed). Prentice Hall, Inc.