Countability & Separation Axioms¶
Separation axioms¶
Separation axioms
Suppose \(X\) is a topological space, we have the following four separation axioms.
(i) \(T_1\) separation axiom. \(\forall x\neq y\in X\), there exsits a neighborhood \(U\) of \(x\) such that \(y\not\in U\).
(ii) \(T_2\) separation axiom. \(\forall x\neq y\in X\), there exsit neighborhood \(U\), \(V\) of \(x\), \(y\), respectively, such that \(U\cap V=\varnothing\).
(iii) \(T_3\) separation axiom (regular). \(\forall x\in X\) and closed set \(A\subset X\), there exist neighborhood \(U\), \(V\) of \(x\) and \(A\) such that \(U\cap V=\varnothing\).
(iv) \(T_4\) separation axiom (normal). For every two disjoint closed set \(A,B\subset X\), there exist neighborhood \(U\), \(V\) of \(A\) and \(B\) such that \(U\cap V=\varnothing\).
A space that satisfies \(T_2\) axiom is called a Hausdorff space, which turned out to have good properties for analysis.
Relationship of serapation axioms
(i) \(T_2\) implies \(T_1\).
(ii) If \(T_1\) axiom holds, then \(T_4\) implies \(T_3\), \(T_3\) implies \(T_2\).
Example. Show that \(X\) is \(T_1\), iff finite-number subset of \(X\) is closed.
Now we give the definition of convergence.
Definition of convergence
Suppose \(X\) is a topological space, with \(\{x_n\}\subset X\) is a sequence, and \(x\in X\). \(x_n\) is said to converge to \(x\), if for all open neighborhood \(U\) of \(x\), there exists \(N>0\), such that \(x_n\in U\) whenever \(n>N\).
Example. If \(X\) is a Hausdorff space, then any sequence in \(X\) converges to at most one point.
Countability axioms¶
Countability axioms
Suppose \(X\) is a topological space, we have the following two countability axioms.
(i) \(C_1\) countability axiom. \(\forall x\in X\), there exsits a countable collection \(\mathcal{B}\) of neighborhoods of \(x\), such that every neighborhood of \(x\) contains some \(B\in \mathcal{B}\).
We also call \(X\) has a countable basis at \(x\).
(ii) \(C_2\) countability axiom. \(X\) has a countable basis for its topology.
\(C_2\) axiom is stronger than \(C_1\), and some metric space could not satisfy the condition.
Example. In \(\mathbb{R}\), define metric
Show that every point of the above space is its open set and contains uncountable basis for its topology.
Choose open ball
which is the smallest element of basis for its topology, and thus is uncountable.
Example. Separable metric space is \(C_2\). Thus \(\mathbb{E}^n\), Hilbert space \(\mathbb{E}^\omega\) are \(C_2\).
Lindelöf Theorem
If a topological space satisfies \(C_2\) and \(T_3\), then it is also \(T_4\). Or in some terms, A regular space with countable basis for its topology is normal.
Choose a countable basis for its topology \(\mathcal{B}\). Suppose \(F\) and \(F'\) are disjoint closed subsets.
Since the space is \(T_3\), so for each \(x\in F\), there exist open sets \(U_x\) \(V_x\) such that \(x\in U_x\), \(F'\subset V_x\) and \(U_x\cap V_x=\varnothing\). Thus we have \(\overline{U_x}\cap F'=\varnothing\). For each \(U_x\), choose an open set \(B_x\in \mathcal{B}\) such that \(B_x\subset U_x\) and thus \(\overline{B_x}\cap F'=\varnothing\). Since \(\mathcal{B}\) is countable, we have
such that for each \(n\), \(\overline{B_n}\cap F'=\varnothing\) and \(\overline{B_n}'\cap F=\varnothing\), and \(F\subset \bigcup_{n} B_n\) and \(F'\subset \bigcup_n B_n'\).
Define open sets
So we have \(U_n\cap V_m=\varnothing\) for all \(n,m\geq 1\). Let
so
Easy to show that \(U\) and \(V\) are the desired open sets that separate \(F\) and \(F'\).
Urysohn Lemma & its applications¶
Urysohn Lemma
Assume topological space \(X\) is \(T_4\), then for any two disjoint closed sets \(A\) and \(B\), there exists a continuous function defined on \(X\) such that \(f|_A=0\) and \(f|_B=1\).