Topology/Countability

From Wikibooks, the open-content textbooks collection

Current revision (unreviewed)

[edit] Bijection

A set is said to be countable it there exists a one to one correspondence between that set and the set of integers.

[edit] Examples

The Even Integers: There is a simple bijection between the integers and the even integers, namely $f:\bold{Z}\rightarrow\bold{Z}$ , where $f (n) = 2 n$ . Hence the even integers are countable.

A 2 - Dimensional Lattice: Let $\bold{Z}^2$ represent the usual two dimensional integer lattice, then $\bold{Z}^2$ is countable.

Proof: let $f:\bold{Z}\rightarrow\bold{Z}$ represent the function such that $f (0) = (0,)$ and $f (n) = (x, y)$ , where $(x, y)$ is whichever point:

not represented by some $f (m)$ for $m < n$

$(x, y)$ is the lattice point 1 unit from $f (n - 1)$ nearest to the origin. In the case where there are two such points, an arbitrary choice may be made.

Because $f$ exists and is a bijection with the integers, The 2 - dimensional integer lattice is countable.

[edit] Axioms of countability

[edit] First Axiom of Countability

[edit] Definition

A topological space $X$ is said to satisfy the First Axiom of Countability if, for every $x\in X$ there exists a countable collection $\mathcal{U}$ of neighbourhoods of $x$ , such that if $N$ is any neighbourhood of $x$ , there exists $U\in\mathcal{U}$ with $U\subseteq N$ .

A topological space that satisfies the first axiom of countability is said to be First-Countable.

All metric spaces satisfy the first axiom of countability because for any neighborhood $N$ of a point $x$ , there is an open ball $B r (x)$ within $N$ , and the countable collection of neighborhoods of $x$ that are $B 1 / k (x)$ where $k\in\mathbb{N}$ , has the neighborhood $B 1 / n (x)$ where $\tfrac{1}{n}<r$ .

[edit] Theorem

If a topological space satisfies the first axiom of countability, then for any point of closure $x$ of a set $S$ , there is a sequence ${a i}$ of points within $S$ which converges to $x$ .

[edit] Proof

Let ${A i}$ be a countable collection of neighborhoods of $x$ such that for any neighborhood $N$ of $x$ , there is an $A i$ such that $A_i\subset N$ . Define
$B_n=\bigcap_{i=1}^{n} A_n$ .

Then form a sequence ${a i}$ such that $a_i\in B_i$ . Then obviously ${a i}$ converges to $x$ .

[edit] Theorem

Let $X$ be a topological space satisfying the first axiom of countability. Then, a subset $A$ of $X$ is closed if an only if all convergent sequences $\{ x_n\}\subset A$ which converge to an element of $X$ converge to an element of $A$ .

[edit] Proof

Suppose that ${x n}$ converges to $x$ within $X$ . The point $x$ is a limit point of ${x n}$ and thus is a limit point of $A$ , and since $A$ is closed, it is contained within $A$ . Conversely, suppose that all convergent sequences within $A$ converge to an element within $A$ , and let $x$ be any point of contact for $A$ . Then by the theorem above, there is a sequence ${x n}$ which converges to $x$ , and so $x$ is within $A$ . Thus, $A$ is closed.

[edit] Theorem

If a topological space $X$ satisfies the first axiom of countability, then $f:X\to Y$ is continuous if and only if whenever ${x n}$ converges to $x$ , ${f (x n)}$ converges to $f (x)$ .

[edit] Proof

Let $X$ satisfy the first axiom of countability, and let $f:X\to Y$ be continuous. Let ${x n}$ be a sequence which converges to $x$ . Let $B$ be any open neighborhood of $f (x)$ . As $f$ is continuous, there exists an open neighbourhood $A\subset f^{-1}(B)$ of $x$ . Since ${x n}$ to $x$ , then there must exist an $N\in\mathbb{N}$ such that $A$ must contain $x n$ when $n > N$ . Thus, $f (A)$ is a subset of $B$ which contains $f (x n)$ when $n > N$ . Thus, ${f (x n)}$ converges to $f (x)$ .
Conversely, suppose that whenever ${x n}$ converges to $x$ , that ${f (x n)}$ converges to $f (x)$ . Let $B$ be a closed subset of $Y$ . Let $x_n\in f^{-1}(B)$ be a sequence which converges onto a limit $x$ . Then $f (x n)$ converges onto a limit $f (x)$ , which is within $B$ . Thus, $x$ is within $f - 1 (B)$ , implying that it is closed. Thus, $f$ is continuous.

[edit] Second Axiom of Countability

[edit] Definition

A topological space is said to satisfy the second axiom of countability if it has a countable base.

A topological space that satisfies the second axiom of countability is said to be Second-Countable.

A topological space satisfies the second axiom of countable is first countable, since the countable collection of neighborhoods of a point can be all neighborhoods of the point within the countable base, so that any neighborhood $N$ of that point must contain at least one neighborhood $A$ within the collection, and $A$ must be a subset of $N$ .

[edit] Theorem

If a topological space $X$ satisfies the second axiom of countability, then all open covers of $X$ has a countable subcover.

[edit] Proof

Let $\mathcal{G}$ be an open cover of $X$ , and let $\mathcal{B}$ be a countable base for $X$ . $\mathcal{B}$ covers $X$ . For all points $x$ , select an element of $\mathcal{G}$ , $C x$ which contains $x$ , and an element of the base, $B x$ which contains $x$ and is a subset of $C x$ (which is possible because $\mathcal{B}$ is a base). ${B x}$ forms a countable open cover for $X$ . For each $B x$ , select an element of $\mathcal{G}$ which contains $B x$ , and this is a countable subcover of $\mathcal{G}$ .

[edit] Seperable Spaces

[edit] Definition

A topological space $X$ is separable if it has a countable proper subset $A$ such that $Cl(A) = X$ .

Example: $\mathbb{R}^n$ is separable because $\mathbb{Q}^n$ is a countable subset and $\mathrm{Cl}(\mathbb{Q}^n)=\mathbb{R}^n$ .

[edit] Theorem

If a topological space satisfies the second axiom of countability, then it is separable.

[edit] Proof

Consider a countable base of a space $X$ . Choose a point from each set within the base. The resulting set $A$ of the chosen points is countable. Moreover, its closure is the whole space $X$ since any neighborhood of any element of $X$ must be a union of the bases, and thus must contain at least one element within the base, which in turn must contain an element of $A$ because $A$ contains at least one point from each base. Thus it is separable.

[edit] Theorem

If a metric space is separable, then it satisfies the second axiom of countability.

[edit] Proof

Let $X$ be a metric space, and let $A$ be a countable set such that $Cl(A) = X$ . Consider the countable set $B$ of open balls $\{ B_{1/k}(p)|k\in N, p\in A\}$ . Let $O$ be any open set, and let $x$ be any element of $O$ , and let $N$ be an open ball of $x$ within $O$ with radius r. Let $r'$ be a number of the form $1 / n$ that is less than $r$ . Because $Cl(A) = X$ , there is an element $x'\in A$ such that $d(x',x)<\tfrac{r'}{4}$ . Then the ball $B r' / 2 (x')$ is within $B$ and is a subset of $O$ because if $y\in B_{r'/2}(x')$ , then $d(y,x)\leq d(y,x')+d(x',x)<\tfrac{3}{4}r'<r$ . Thus $B_{r'/2}\subseteq O$ that contains $x$ . The union of all such neighborhoods containing an element of $O$ is $O$ . Thus $B$ is a base for $X$ .

[edit] Corollary (Lindelöf covering theorem)

If a metric space is separable, then it satisfies the second axiom of countability, and thus any cover of a subset of that metric space can be reduced to a countable cover.

Example: Since $\mathbb{R}^n$ is a separable metric space, it satisfies the second axiom of countability. This directly implies that any cover a set in $\mathbb{R}^n$ has a countable subcover.

[edit] Countable Compactness

[edit] Definition

A subset $A$ of a topological space $X$ is said to be Countably Compact if and only if all countable covers of $A$ have a finite subcover.

Clearly all compact spaces are countably compact.

A countably compact space is compact if it satisfies the second axiom of countability by the theorem above.

[edit] Theorem

A topological space $X$ is countably compact if and only if any infinite subset of that space has at least one limit point.

[edit] Proof

( $\Rightarrow$ )Let ${x i}$ , $(i = 1,2,3,...)$ be a set within $X$ without any limit point. Then this sequence is closed, since they are all isolated points within the sequence. Let $S n = {x i}$ for $(i = n, n + 1, n + 2,...)$ . The $X\setminus S_n$ are all open sets, and so is a countable cover of the set, but any finite subcover $\{ X\setminus S_{n_i}\}$ of this cover does not cover $X$ because it does not contain $S_{n_{max\{ i\} }}$ . This contradicts the assumption that $X$ is countably compact.

( $\Leftarrow$ )Let ${S n}$ be open subsets of $X$ such that any finite union of those sets does not cover $X$ . Define:

$B_n=\bigcup_{i=1}^{n} S_n$ ,

which does not cover $X$ , and is open. Select $x n$ such that $x_n\notin B_n$ . There is a limit point $x$ of this set of points, which must also be a limit point of $X\setminus B_n$ . Since $X\setminus B_n$ is closed, $x\in X \setminus B_n$ . Thus, $x\notin B_n$ and thus is not within any $S n$ , so $S n$ is not an open cover of X. Thus, $X$ is countably compact.

[edit] Relative Countable Compactness

Since there is relative compactness, there is an analogous property called relative countable compactness.

[edit] Definition

A subset S of a topological space X is relatively countably compact when its closure Cl(S) is countably compact.

[edit] Total Boundedness

[edit] Definition

A set $N\subseteq X$ is an $\varepsilon$ -net of a metric space $X$ where $\varepsilon>0$ if for any $b$ within $X$ , there is an element $x\in N$ such that $d(b,x)<\varepsilon$ .

[edit] Definition

A metric space $X$ is totally bounded when it has a finite $\varepsilon$ -net for any $\varepsilon>0$ .

[edit] Theorem

A countably compact metric space is totally bounded.

[edit] Proof

Any infinite subset of a countably compact metric space $X$ must have at least one limit point. Thus, selecting $x_1,x_2,x_3,\ldots$ where $x n$ is at least $\varepsilon$ apart from any $x d$ where $d < n$ , one must eventually have formed an $\varepsilon$ -net because this process must be finite, because there is no possible infinite set with all elements more than $\varepsilon$ apart.

[edit] Theorem

A totally bounded set is separable.

[edit] Proof

Take the union of all finite $1 / n$ -nets, where $n$ varies over the natural numbers, and that is a countable set such that its closure is the whole space $X$ .

[edit] Urysohn's Metrizability Theorem

The following theorem establishes a sufficient condition for a topological space to be metrizable.

[edit] Theorem

A second countable normal T1 topological space is homeomorphic to a metric space.

[edit] Proof

We are going to use the Hilbert cube, which is a metric space, in this proof, to prove that the topological space is homeomorphic to a subset of the Hilbert cube, and is thus a metric space.

First, since all T1 normal spaces are Hausdorff, all single points are closed sets. Therefore, consider any countable base of the topological space X, and any open $O n$ set of it. Select a point $x n$ within this open set. Since the complement of the open set is closed, and since a point within the open set is also closed, and since these two closed sets are disjoint, we can apply Urysohn's lemma to find a continuous function $f_n: X \rightarrow [0,1]$ such that:

$f n (x n) = 0$
$f n (X / O n) = 1$

It is easy to see from the proof of Urysohn's lemma that we have not only constructed a function with such properties, but that such that $f n (O n) < 1$ , meaning that the function value of any point within the open set is less than 1.

Now define the function $g: X \rightarrow H$ from X to the Hilbert cube to be $g(x)=(f_1(x),\frac{f_2(x)}{2},\frac{f_3(x)}{4},...)$ .

To prove that this is continuous, let $a_n \rightarrow a$ be a sequence that converges to a. Consider the open ball $B ε (f (a))$ where $ε > 0$ . There exists an N such that

$\sum_{i=N}^\infty(\frac{1}{2^i})^2<\frac{\epsilon^2}{2}$ .

Moreover, since $f n$ is a continuous function from X to [0,1], there exists a neighborhood of $a$ , and therefore an open set $S n$ of the base within that neighborhood containing a such that if $y\in S_n$ , then

$|f_n(y)-f_n(z)|<\frac{2^n\epsilon}{\sqrt{2N}}$

or

$(\frac{f_n(y)-f_n(z)}{2^n})^2<\frac{\epsilon^2}{2N}$ .

Let

$S=\bigcap_{i=1}^{N-1} S_i$ .

In addition, since $a_n\rightarrow a$ , there exists an $M i$ (i=1,2,3,...,M-1) such that when $n > M i$ , that $a_n\in S_i$ , and let M be the maximum of $M i$ so that when n>M, then $a_n\in S$ .

Let n>M, and then the distance from $g (a n)$ to g(a) is now

$\sum_{i=1}^\infty (\frac{f_i(a_n)-f_i(a)}{2^i})^2 =$ $\sum_{i=1}^{N-1}(\frac{f_i(a_n)-f_i(a)}{2^i})^2 +$ $\sum_{i=N}^\infty(\frac{f_i(a_n)-f_i(a)}{2^i})^2 \le$ $\frac{N-1}{2N}\epsilon^2+\sum_{i=N}^\infty(\frac{f_n(y)-f_n(z)}{2^n})^2 \le$ $\frac{N-1}{2N}\epsilon^2+\frac{\epsilon^2}{2} < \epsilon^2.$

This proves that it is continuous.

To prove that this is one-to-one, consider two different points, a and b. Since the space is Hausdorff, there exists disjoint open sets $a\in U_a$ and $b\in U_b$ , and select an element of the base $O n$ that contains a and is within $U a$ . It follows that $f n (a) < 1$ whereas $f n (b) = 1$ , proving that the function g is one-to-one, and that there exists an inverse $g - 1$ .

To prove that the inverse $g - 1$ is continuous, let $O n$ be an open set within the countable base of X. Consider any point x within $O n$ . Since $f n (x) < 1$ , indicating that there exists an $ε n > 0$ such that when

$| f n (z) - f n (x) | < 2 n ε n$

then $z\in O_n$ .

Suppose that $g(z)\in g(X)\cap B_{\epsilon_n}(g(y))$ . Then

$(\frac{f_n(z)-f_n(x)}{2^n})^2\le\sum_{i=1}^\infty(\frac{f_i(z)-f_i(x)}{2^i})^2\le \epsilon_n^2$

Implying that $|f_n(z)-f_n(x)|\le 2^n\epsilon^n$ indicating that $z\in O_n$ .

Now consider any open set O around x. Then there exists a set of the base $x\in O_n\subseteq O$ and an $ε n > 0$ such that whenever $g(z)\in g(X)\cap B_{\epsilon_n}(g(y))$ , then $z\in O_n$ , meaning that $z\in O$ . This proves that the inverse is continuous.

Since the function is continuous, is one-to-one, and has a continuous inverse, it is thus a homeomorphism, proving that X is metrizable.

Note that this also proves that the Hilbert cube thus contains any second-countable normal T1 space.

[edit] Hahn-Mazurkiewicz Theorem

The Hilbert Curve- a space filling curve

The Hahn-Mazurkiewicz theorem is one of the most historically important results of point-set topology, for it completely solves the problem of "space-filling" curves. This theorem provides the necessary and sufficient condition for a space to be 'covered by curve', a property that is widely considered to be counter-intuitive.

Here, we present the theorem without its proof.

[edit] Theorem

A Hausdorff space is a continuous image of the unit interval $[0,1]$ if and only if it is a compact, connected, locally connected and second-countable space.

[edit] Exercise

Prove that a separable metric space satisfies the second axiom of countability. Hence, or otherwise, prove that a countably compact metric space is compact.
Prove the sufficiency condition of the Hahn-Mazurkiewicz theorem:
If a Hausdorff space is a continuous image of the unit interval, then it is compact, connected, locally-connected and second countable.

Topology/Countability

From Wikibooks, the open-content textbooks collection

Contents

[edit] Bijection

[edit] Examples

[edit] Axioms of countability

[edit] First Axiom of Countability

[edit] Definition

[edit] Theorem

[edit] Proof

[edit] Theorem

[edit] Proof

[edit] Theorem

[edit] Proof

[edit] Second Axiom of Countability

[edit] Definition

[edit] Theorem

[edit] Proof

[edit] Seperable Spaces

[edit] Definition

[edit] Theorem

[edit] Proof

[edit] Theorem

[edit] Proof

[edit] Corollary (Lindelöf covering theorem)

[edit] Countable Compactness

[edit] Definition

[edit] Theorem

[edit] Proof

[edit] Relative Countable Compactness

[edit] Definition

[edit] Total Boundedness

[edit] Definition

[edit] Definition

[edit] Theorem

[edit] Proof

[edit] Theorem

[edit] Proof

[edit] Urysohn's Metrizability Theorem

[edit] Theorem

[edit] Proof

[edit] Hahn-Mazurkiewicz Theorem

[edit] Theorem

[edit] Exercise

Views

Personal tools

Navigation

Search

community

Print/export

Toolbox