5.4 KiB
Math4201 Topology I (Lecture 28)
Compact spaces
Extreme value theorem
Definition of diameter
Let (X,d) be a metric space and A\subseteq X. The diameter of A is defined as
\operatorname{diam}(A) = \sup\{d(x,y):x,y\in A\}
Lebesgue number lemma
Let X be a compact metric space and \{U_\alpha\}_{\alpha\in I} be an open cover of X. Then there is \delta>0 such that for every subset A\subseteq X with diameter less than \delta, there is \alpha\in I such that A\subseteq U_\alpha.
Proof
Consider x\in X, there is an element U_\alpha in the open covering such that x\in U_\alpha.
In particular, there is r_x such that B_{r_x}(x)\subseteq U_\alpha.
Then the collection \{B_{\frac{r_x}{2}}(x)\}_{x\in X} is an open covering of X. (each x\in X is contained in some B_{\frac{r_x}{2}}(x))
Since X is compact, there is a finite subcover \{B_{\frac{r_{x_i}}{2}}(x_i)\}_{i=1}^n of X. Such that \bigcup_{i=1}^n B_{\frac{r_{x_i}}{2}}(x_i)=X.
Let \delta = \min\{r_{\frac{r_{x_1}}{2}}, ..., r_{\frac{r_{x_n}}{2}}\}>0.
Let A\subseteq X be a subset with diameter less than \delta.
Take y\in A, then A\subseteq B_\delta(y).
Take x_i such that y\in B_{\frac{r_{x_i}}{2}}(x_i). (such cover exists by definition of the subcover)
And then \alpha such that B_{\frac{r_{x_i}}{2}}(x_i)\subseteq U_\alpha.
We claim that B_\delta(y)\subseteq U_\alpha, which would imply that A\subseteq U_\alpha.
y\in B_{\frac{r_{x_i}}{2}}(x_i), and we know that B_{r_{x_i}}(x_i)\subseteq U_\alpha.
Since \delta < \frac{r_{x_i}}{2}, it suffices to show that B_{\frac{r_{x_i}}{2}}(y)\subseteq U_\alpha.
For any z\in B_{\frac{r_{x_i}}{2}}(y), we have d(z,y)<\frac{r_{x_i}}{2}, and d(y,x_i)<\frac{r_{x_i}}{2}.
So d(z,x_i)\leq d(z,y)+d(y,x_i)<\frac{r_{x_i}}{2}+\frac{r_{x_i}}{2}=r_{x_i}, so z\in B_{r_{x_i}}(x_i)\subseteq U_\alpha.
So B_{\frac{r_{x_i}}{2}}(y)\subseteq U_\alpha.
Definition of finite intersection property
A collection \{C_\alpha\}_{\alpha\in I} of subsets of a set X has finite intersection property if for every finite subcollection \{C_{\alpha_1}, ..., C_{\alpha_n}\} of \{C_\alpha\}_{\alpha\in I}, we have \bigcap_{i=1}^n C_{\alpha_i}\neq \emptyset.
Theorem
A space X is compact if and only if every collection \{Z_\alpha\}_{\alpha\in I} of closed subsets of X satisfies the finite intersection property has a non-empty intersection.
\bigcap_{\alpha\in I} Z_\alpha \neq \emptyset
Non-example
Consider X=(0,1) is not compact with the standard topology.
Consider Z_n=(0,\frac{1}{n}], each interval is closed in X. This satisfies the finite intersection property because \bigcap_{i=1}^k Z_{n_i}\neq \emptyset for any finite subcollection \{Z_{n_1}, ..., Z_{n_k}\}. We can find a smaller for any finite subcollection to get a non-empty intersection.
But \bigcap_{n=1}^\infty Z_n = \emptyset.
Proof
\implies
Let U_\alpha=X-Z_\alpha is open for each \alpha\in I. By contradiction, suppose that \bigcap_{\alpha\in I} Z_\alpha = \emptyset.
So X-\bigcap_{\alpha\in I} Z_\alpha = X=\bigcup_{\alpha\in I} U_\alpha = \bigcup_{\alpha\in I} (X-Z_\alpha)=X.
So \{U_\alpha\}_{\alpha\in I} is an open cover of X. Since X is compact, there is a finite subcover \{U_{\alpha_1}, ..., U_{\alpha_n}\}.
So \bigcap_{i=1}^n U_{\alpha_i} = X-\bigcup_{i=1}^n Z_{\alpha_i} = X, \bigcap_{i=1}^n Z_{\alpha_i} = \emptyset. This contradicts the finite intersection property.
\impliedby
Proof is similar.
Definition of isolated point
A point x\in X is an isolated point if \{x\} is an open subset of X.
Example of isolated point
X=[0,1]\cup \{2\} with subspace topology from \mathbb{R}.
Then \{2\} is an isolated point \{2\}=X\cap (2-\frac{1}{2}, 2+\frac{1}{2}).
Theorem of compact Hausdorff spaces without isolated points
Any non-empty compact Hausdorff space without an isolated point is uncountable.
Proof
Proof by contradiction.
Let X=\{x_n\}_{n\in\mathbb{N}} be a countable set.
Since x_1 is not an isolated point, so there exists y_1\in X such that y_1\neq x_1. Apply the Hausdorff property, there exists disjoint open neighborhoods U_1 and V_1 such that x_1\in U_1 and y_1\in V_1.
In particular \overline{V_1} does not contain x_1, but it contains y_1. (Follows from disjoint open neighborhoods)
Since x_2 is not an isolated point, so there exists y_2\in X such that y_2\neq x_2. Apply the Hausdorff property, there exists disjoint open neighborhoods U_2 and V_2 such that x_2\in U_2 and y_2\in V_2.
If x_2\notin V_1, then we define V_2 as V_1.
If x_2\in V_1, then by the assumption, there is another point y_2 in V_1 which isn't the same as x_2.
CONTINUE NEXT TIME.
Theorem real numbers is uncountable
\mathbb{R} is uncountable, and any interval in \mathbb{R} is uncountable.
Proof
It suffices to prove this for a closed interval [a,b] with a<b. Because any interval contains such a closed interval.
The claim for a closed interval [a,b] follows from the following theorem because [a,b] is a non-empty compact Hausdorff space without an isolated point.