Subspaces and products

Properties

• If $X$ is metrizable and $Y$ is a subspace of $X$, then $Y$ is metrizable.
• If $X$ is metrizable, then $X$ is Hausdorff.
• If $(X_{i})_{i\in I}$ is a countable family of metrizable spaces, then $\prod_{i\in I} X_{i}$ is metrizable.

Continuous functions

1. $\epsilon$ - $\delta$ definition

   Let $f\colon X\to Y$ a function, with $X$, $Y$ metrizable spaces with metrics $d_{X}$, $d_{Y}$, respectively. Then $f$ is continuous if and only if for all $x\in X$, $\epsilon>0$ there is $\delta>0$ such that

   $$d_{Y}(f(x),f(y))<\epsilon \text{ if } d_{X}(x,y)<\delta$$

Sequences

Definition

1. Sequence

   If $X$ is a topological space, a sequence on $X$ is a function $\mathbb{N}\to X$.

2. Notation

   We denote a sequence $(x_n)$ or $(x_1,x_2,\ldots)$.

3. Convergence

   We say that the sequence $(x_n)$ converges to $x\in X$ if for every neighborhood $U$ of $x$ there is a positive integer $N$ such that $x_n\in U$ for all $n\geq N$. We write $x_n\to x$.

Remarks on the definition of convergence

• A sequence does not necessarily converges, and if it does, the limit is not necessarily unique.
• If $X$ is Hausdorff, a sequence on $X$ that converges has a unique limit.

Sequence lemma

1. Sequence lemma

   Let $X$ be a topological space, and $A\subseteq X$. If $(x_n)$ is a sequence in $A$ that converges to $x$, then $x\in\overline{A}$.

   Conversely, if $x\in\overline{A}$ and $X$ es metrizable, then there is a sequence on $A$ that converges to $x$

Proof of the sequence lemma

1. Proof

   • Suppose that $x_n\to x$, where $x_{n}\in A$ for all $n$. Let $U$ be a neighborhood of $x$. By definition of convergence, $x_n\in U$ for big enough values of $n$, and so $x\in\overline{A}$.

   • Now, let $X$ be metrizable with metric $d$, and let $x\in\overline{A}$. For each $n$, we choose a point $x_{n}\in A\cap B(x,\frac{1}{n})$.

   • We claim that $x_n\to x$. Because if $U$ is a neighborhood of $x$, let $\epsilon>0$ be such that $B(x,\epsilon)\subseteq U$, and let $N>\frac{1}{\epsilon}$.

   • Then, if $n\geq N$, we have that $\frac{1}{n}\leq \frac{1}{N}<\epsilon$, and so $x_{n}\in U$ si $n\geq N$.

Remarks to the proof

• The theorem gives a necessary condition for metrizability. We will see examples of $A\subseteq X$, $x\in X$ with $x\in\overline{A}$ such that there is no sequence in $A$ that converges to $x$. In such cases, $X$ is not metrizable.
• In the proof of the reciprocal, we do not need the full hypothesis of metrizability, but the properties of the collection of balls $\{B(x,\frac{1}{n})\}_{n\in\mathbb{N}}$.
• A countable local base on $x\in X$ is a collection $\{B_{n}\}_{n\in\mathbb{N}}$ of neighborhoods of $x$ such that for any neighborhood $U$ of $x$ there is $n$ such that $B_{n}\subseteq U$.
• The topological space $X$ is first–countable if every $x\in X$ has a countable local base.
• (Exercise) Prove that if $X$ es first–countable, $A\subseteq X$ and $x\in\overline{A}$, there is a sequence $(x_n)$ on $A$ that converges to $x$.

Continuity and sequences

1. Continuity by sequences

   Let $f\colon X\to Y$ with $X$ metrizable. Then $f$ is continuous if and only if for all convergent sequences $x_{n}\to x$ on $X$, the sequence $f(x_n)$ converges to $f(x)$.

2. Proof

   • Suppose $f$ continuous and let $x_n\to x$. We want to prove $f(x_{n})\to f(x)$. Let $V$ be a neighborhood of $f(x)$. Then $f^{-1}(V)$ is a neighborhood of $x$, and so there is $N$ such that $x_{n}\in f^{-1}(V)$ if $n\geq N$. But then $f(x_{n})\in V$ for such values of $n$.

Continuation of the proof

1. Continuation of the proof
   • For the converse, suppose that for any convergent sequence $x_{n}\to x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$. We want to prove that $f$ is continuous.
   • Let $A\subseteq X$, we want to see that $f(\overline{A})\subseteq \overline{f(A)}$.
   • Let $x\in\overline{A}$. Since $X$ is metrizable, there is a sequence $x_n$ on $A$ such that $x_{n}\to x$.
   • By hypothesis, we have that $f(x_{n})\to f(x)$. Since $f(x_{n})\in f(A)$, then $f(x)\in\overline{f(A)}$. Hence $f(\overline{A})\subseteq \overline{f(A)}$.

Examples

Box topology on $\mathbb{R}^{\mathbb{N}}$

• We will see that the sequence lemma does not hold in $X=\mathbb{R}^{\mathbb{N}}$ with the box topology.

• Let $A$ be the set of all real sequences where the terms are all positive:

  $$A=\{(x_1,x_2,\ldots)\mid x_i>0\text{ for all \(i\)}\}$$

• Let $0=(0,0,\ldots)\in X$. Then $0\in\overline{A}$, since if

  $$B=(a_1,b_1)\times(a_2,b_2)\times\cdots$$

  is a basic subset that contains $0$, then $B\cap A\ne\emptyset$.

Continuation

• However, there is no sequence on $A$ that converges to $0$. Suppose that $(a_n)$ is a sequence on $A$. Let

  $$a_{n}=(x_{1n},x_{2n},\ldots,x_{in},\ldots)$$

• Since each $x_{in}$ is greater than zero, the basic subset of $X$:

  $$B'=(-x_{11},x_{11})\times(-x_{22},x_{22})\times\cdots$$

  is a neighborhood of $0\in X$.

• However, $B'$ does not contain points of the sequence $(a_n)$. Hence $(a_n)$ does not converge to $0$.

Uncountable product of copies of $\mathbb{R}$

• Let $J$ be uncountable. We show that $\mathbb{R}^{J}$ with product topology is not metrizable.
• Let $A\subseteq \mathbb{R}^{J}$, where $A$ consists of the $(x_{\alpha})$ such that $x_{\alpha}=1$ for all but a finite number of $\alpha\in J$, where we have $x_{\alpha}=0$. Let $0$ denote the “origin” of $\mathbb{R}^{J}$.
• We prove that $0\in \overline{A}$. Let $\prod U_{\alpha}$ be a basis element such that $0\in \prod U_{\alpha}$. Suppose $U_{\alpha}\ne \mathbb{R}$ for $\alpha\in\{\alpha_{1},\ldots,\alpha_{n}\}$.
• Then, if $(x_{\alpha})$ is defined as $x_{\alpha}=0$ for $\alpha\in\{\alpha_{1},\ldots,\alpha_{n}\}$ and $x_{\alpha}=1$ for all other values of $\alpha$, we have that $(x_{\alpha})\in A\cap \prod U_{\alpha}$.
• We will show then that there is no sequence of points in $A$ that converges to $0$. Let $(a_{n})$ a sequence in $A$.

Continuation

• Each $a_{n}\in A$ has only a finite number of values of $\alpha$ such that the $\alpha$th term of $a_{n}$ is equal to $0$. Let $J_{n}$ be the set of such values of $\alpha$.
• The union of all the $J_{n}$ is a countable union of finite sets, hence countable. Since $J$ is uncountable, let $\beta\in J$ that is not in any of the $J_{n}$. That means that the $\beta$th coordinate of all of the $a_{n}$ is 1.
• Let $U_{\beta}=(-1,1)$, and let $U=\pi_{\beta}^{-1}(U_{\beta})$. Then $U$ is a neighborhood of $0\in \mathbb{R}^{J}$, and $U$ does not contain any of the $a_{n}$. Hence $a_{n}$ does not converge to $0$.

Well-ordered sets

Definition and examples

1. Well-orderet set

   We say that the totally ordered set $(X,\leq)$ is well-ordered if every nonempty subset has a minimum element.

2. Examples

   • The set $\mathbb{N}$ with the usual order is well-ordered.
   • The set $\mathbb{R}$ with the usual order is not well-ordered.

Properties

1. Properties of well-ordered sets
   • If $X$ is well-ordered and $Y\subseteq X$ has the induced order, then $Y$ is well-ordered.
   • If $X,Y$ are well-ordered sets, then $X\times Y$ with dictionary order is well-ordered.

Zermelo’s theorem

1. Zermelo

   If $A$ is any set, there is a total order $\leq$ on $A$ that is a well-order.

2. Corolario

   There is a set that is uncountable and well-ordered.

Sections

1. Section

   If $S$ is an ordered set and $\alpha\in S$, we denote by $S_{\alpha}$ the set:

   $$S_{\alpha}=\{x\in S\mid x<\alpha\}.$$

   $S_{\alpha}$ is called the section of $S$ by $\alpha$.

The example

1. Teorema

   There is a set that is uncountable and well-ordered, but every section of it is countable.

2. Proof

   • Let $X$ be an uncountable, well-ordered set. Then the set $S=\{1,2\}\times X$ is also uncountable and well-ordered, and has sections that are uncountable (for example, those of elements of the form $(2,x)$.)

   • The set

     $$\{\alpha\in S\mid \text{\(S_{\alpha}\) is uncountable}\}$$

     is nonempty, let $\Omega$ its smallest element.

   • Then $S_{\Omega}$ is the required set.

A corollary

1. Corolario

   If $A\subseteq S_{\Omega}$ is countable, then $A$ has an upper bound in $S_{\Omega}$.

2. Proof
   • Since for each $a\in A$, we have that $S_{a}$ is countable, then $B=\cup_{a\in A}S_{a}$ is also countable.
   • Since $B\subseteq S_{\Omega}$, and $S_{\Omega}$ is uncountable, we can choose $x\in S_{\Omega}-B$.
   • If there was $a\in A$ with $x<a$, then $x$ would be in $B$, which is a contradiction.
   • Hence $x$ is an upper bound of $A$.\qed

3. A nonmetrizable ordered space

   The set $S_{\Omega}\cup\{\Omega\}$ is not metrizable

4. Proof
   • We have that $\Omega$ is a limit point of $S_{\Omega}$, since any basic element containing $\Omega$ has the form $(a,\Omega]$ and must intersect $S_{\Omega}$. Otherwise we would have $S_{\Omega}=S_{a}\cup\{a\}$, but $S_{a}\cup\{a\}$ is countable.
   • However, there is no sequence in $S_{\Omega}$ that converges to $\Omega$, since if $(x_{n})$ is a sequence in $S_{\Omega}$, there is an upper bound $b\in S_{\Omega}$ for all the terms in the sequence. Then $(b,\Omega]$ contains no point of the sequence.\qed

5. We will denote with $\overline{S_{\Omega}}$ the set $S_{\Omega}\cup\{\Omega\}$.