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\noindent
\parbox{2.25in}{\bf Math 4710/5710}
\hfill {\Large \bf Separation Axioms} \hfill
\parbox{2.25in}{\bf \hfill Fall 2014}

\vspace{0.2in}

We now come to the seedy underbelly of topology. It is hard to believe in this day and age that there are those who still practice radical segregation -- even advocate for it! In the time of Felix Hausdorff (1868--1942) and Pavel Urysohn (1898--1924) this was accepted culturally, but how such separation persists today is hard to understand. You see, topologists love to separate. They have even convinced themselves that it's a ``regular'' or ``normal'' thing. 

Before we can begin to battle these prejudices, we must better understand them. Let's explore this despicable list of axioms in order that we may effect a change and move towards a world which isn't $T_4$ or $T_3$. Maybe our children can grow up not knowing the horrors of Hausdorff ``spaces'' and never having been inflicted with Urysohn's innocent sounding ``lemma''. Stand up for what's right! Be the change you believe in! 

\vspace{0.1in}

Let $X$ be a topological space. We say two points (or more generally two sets) are {\bf topologically distinguishable} if there is an open set containing one but not the other. 

Two sets are {\bf separated} if each set is disjoint from the other set's closure. So $A,B \subseteq X$ are {\bf separated} if $A \cap \overline{B} = \varnothing$ and $\overline{A} \cap B = \varnothing$. [Note: This does not mean that their closures are disjoint!] Notice that separated objects are topologically distinguishable ($B$ is disjoint from $X-\overline{B}$ and $X-\overline{B}$ is an open set containing $A$).

Points (or more generally sets) are {\bf separated by open sets} if the points (or sets) are contained in disjoint open sets. That is, $A, B \subseteq X$ are {\bf separated by open sets} if there exists open sets $U,V \subseteq X$ such that $A \subseteq U$, $B \subseteq V$, and $U \cap V = \varnothing$. As before notice that ``separated by open sets'' implies plain old ``separated''.

Sets are {\bf separated by continuous functions} if there is a continuous function into the reals which takes on the value of 0 at every point in the first set and 1 at every point in the second set. That is, $A, B \subseteq X$ are {\bf separated by a continuous function} $f:X \to [0,1]$ if $f$ is continuous, $f(a)=0$ for all $a \in A$, and $f(b)=1$ for all $b \in B$. Notice ``separated by a continuous function'' implies ``separated by open sets'' since $A \subseteq U = f^{-1}(-1/2,1/2)$ and $B \subseteq V = f^{-1}(1/2,3/2)$ where $U$ and $V$ are open (since $f$ is continuous) and disjoint (because $(-1/2,1/2)$ and $(1/2,3/2)$ are disjoint).  

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\begin{description} \setlength\itemsep{1pt}
\item[Hausdorff] A topological space is called {\bf Hausdorff} if points can be separated by open sets. So for all $x,y \in X$ such that $x \not= y$, there exists open sets $U,V \subseteq X$ such that $x \in U$, $y \in V$, and $U \cap V = \varnothing$.
\item[Regular] A topological space is called {\bf regular} if points and closed sets can be separated by open sets. Specifically, for all $x \in X$ and all closed sets $A \subseteq X$ with $x \not\in A$, there exists open sets $U,V \subseteq X$ such that $x \in U$, $A \subseteq V$, and $U \cap V = \varnothing$.
\item[Completely Regular] A topological space is called {\bf completely regular} if points and closed sets can be separated by continuous functions. Specifically, for all $x \in X$ and all closed sets $A \subseteq X$ with $x \not\in A$ there exists a continuous function $f:X \to [0,1]$ such that $f(x)=0$ and $f(a)=1$ for all $a \in A$.
\item[Normal]  A topological space is called {\bf normal} if closed sets can be separated by open sets. Specifically, for all closed sets $A,B \subseteq X$ with $A \cap B = \varnothing$, there exists open sets $U,V \subseteq X$ such that $A \subseteq U$, $B \subseteq V$, and $U \cap V = \varnothing$. 

{\it Note:} Urysohn's Lemma (see below) tells us that we don't need to define a collection of ``Completely Normal'' spaces. If we did, we would find that completely normal and normal define the same collection of spaces.
\end{description}

\begin{description} \setlength\itemsep{1pt}
\item[\sep{0}] A topological space is said to be \sep{0} if all points are topologically distinguishable. This means that given $x,y \in X$ such that $x \not= y$ there is an open set $U \subseteq X$ such that either $x \in U$ and $y \not\in U$ or $x \not\in U$ and $y \in U$.
\item[\sep{1}] A topological space is said to be \sep{1} if all points are separated. This means that there exists open sets $U,V \subseteq X$ such that $x \in U$ but $y \not\in U$ and $y \in V$ but $x \not\in V$. This is equivalent to saying $x \not\in \overline{\{y\}}$ and $y \not\in \overline{\{x\}}$. In other words, being \sep{1} is equivalent to $\overline{\{x\}}=\{x\}$ for all $x \in X$ -- that is -- {\it singletons are closed}.
\item[\sep{2}] A topological space is said to be \sep{2} if it's Hausdorff: \quad \sep{2} = Hausdorff.
\item[\sep{3}] A topological space is said to be \sep{3} if it's regular and \sep{0}: \quad \sep{3} = regular + \sep{0}.
\item[\sep{3\unitfrac{1}{2}}] A topological space is said to be \sep{3\unitfrac{1}{2}} if it's completely regular and \sep{0}: \quad \sep{3\unitfrac{1}{2}} = completely regular + \sep{0}.
\item[\sep{4}] A topological space is said to be \sep{4} if it's normal and \sep{1}: \quad \sep{4} = normal + \sep{1}.
\end{description}

%\vspace{0.05in}

It can be shown that \sep{4} $\Rightarrow$ \sep{3\unitfrac{1}{2}} $\Rightarrow$ \sep{3} $\Rightarrow$ \sep{2} $\Rightarrow$ \sep{1} $\Rightarrow$ \sep{0}. There are examples which show none of the implications can be reversed. Also, please be careful. What Munkres calls normal, completely regular, and regular are more acurately \sep{4}, \sep{3\unitfrac{1}{2}}, and \sep{3}.

\vspace{0.05in}

\noindent
{\bf Urysohn's Lemma:} [Normal $\Longrightarrow$ Completely Normal] In a normal space, disjoint closed sets can be separated by a continuous function. Specifically, let $X$ be normal with $A,B \subseteq X$ such that $A$ and $B$ are closed and $A \cap B = \varnothing$. Then there exists a continuous function $f:X \to [0,1]$ such that $f(A)=0$ and $f(B)=1$.

\newpage

\noindent
\parbox{2.25in}{\bf Math 4710/5710}
\hfill {\Large \bf Odds \& Ends} \hfill
\parbox{2.25in}{\bf \hfill Fall 2014}

\vspace{0.2in}

\begin{description} \setlength\itemsep{1pt}
\item[First Countable] A topological space $X$ is {\bf first countable} if at each $x \in X$ there is a countable neighborhood basis. This means there is a countable collection $\mathcal{B}$ of neighborhoods of $x$ such that for every neighborhood $U$ of $x$ there exists some $B \in \mathcal{B}$ such that $B \subseteq U$.  [Every metric space is first countable: Consider balls centered at $x \in X$ with positive rational radii.]
\item[Second Countable] A topological space $X$ is {\bf second countable} if it has a countable basis. [$\mathbb{R}^n$ is second countable: Consider balls of rational radii centered at points with rational coordinates.]
\item[Separable] We say $D$ is {\bf dense} in $X$ if $\overline{D} = X$. $X$ is said to be {\bf separable} if it contains a countable dense set. [For example: $\overline{\mathbb{Q}} = \mathbb{R}$ so $\mathbb{R}$ is separable. More generally, any second countable space is separable.]
\item[Lindel\"of] If every open cover has a finite subcover, our space is {\bf compact}. If every {\it countable} open cover (an open cover with only countably many elements) has a finite subcover, then our space is {\bf countably compact}. If every open cover has a {\it countable} subcover, our space is {\bf Lindel\"of}. [Obviously, Lindel\"of + countably compact = compact. Also, every second countable space is Lindel\"of.]
\end{description}

\begin{description} \setlength\itemsep{1pt}
\item[Theorem:] \sep{3} + second countable $\Rightarrow$ \sep{4}.
\item[Theorem:] metrizable $\Rightarrow$ \sep{4}.
\item[Theorem:] compact + Hausdorff $\Rightarrow$ \sep{4}.
\item[Theorem:] Well ordered sets with order topology are \sep{4}.
\item[Urysohn's Metrization Theorem:] \sep{3} + second countable $\Rightarrow$ metrizable.
\item[An Embedding Theorem:] A space is \sep{3\unitfrac{1}{2}} if and only if it is homeomorphic to a subspace of $[0,1]^J$ for some $J$. 
\end{description}

\begin{description} \setlength\itemsep{1pt}
\item[Locally Finite] Let $X$ be a topological space and $\mathcal{A} \subseteq \mathcal{P}(X)$. We say that $\mathcal{A}$ is {\bf locally finite} in $X$ if for every $x \in X$ there exists some neighborhood $U$ of $x$ such that $U$ is disjoint from all but finitely many elements of $\mathcal{A}$. Similarly, $\mathcal{B} \subseteq \mathcal{P}(X)$ is said to be {\bf countably locally finite} (or {\bf $\sigma$-locally finite}) in $X$ if $\mathcal{B} = \bigcup\limits_{n=1}^\infty \mathcal{B}_n$ where each $\mathcal{B}_n$ is locally finite in $X$. This means $\mathcal{B}$ is a countable union of locally finite sets.
\item[Open Refinement] Let $\mathcal{A} \subseteq \mathcal{P}(X)$. We say that $\mathcal{B}$ is an {\bf refinement} of $\mathcal{A}$ if for every $A \in \mathcal{A}$ there is an element $B \in \mathcal{B}$ such that $B \subseteq A$. We say that $\mathcal{B}$ is an {\bf open refinement} of $\mathcal{A}$ if $\mathcal{B}$ is a refinement of $\mathcal{A}$ and every element of $\mathcal{B}$ is an open set. 
\item[Paracompact] A topological space $X$ is {\bf paracompact} if every open cover of $X$ has a locally finite open refinement which covers $X$. 
\item[Locally Metrizable] $X$ is {\bf locally metrizable} if every $x \in X$ has a metrizable neighborhood.
\end{description}

\begin{description} \setlength\itemsep{1pt}
\item[Nagata-Smirnov Metrization Theorem:] A space is metrizable if and only if it is \sep{3} and has a countably locally finite basis.
\item[Theorem:] Hausdorff  + paracompact $\Rightarrow$ \sep{4}.
\item[Theorem:] \sep{3} + Lindel\"of $\Rightarrow$ paracompact.
\item[Smirnov Metrization Theorem:] metrizable $\Leftrightarrow$ Hausdorff + paracompact + locally metrizable. 
\end{description}

\begin{description} \setlength\itemsep{1pt}
\item[Baire Space] $X$ is a {\bf Baire space} if given any countable collection $\mathcal{A} = \{ A_n \;|\; n=1,2,\dots \}$ of closed sets of $X$ such that $\mathrm{int}(A_n)=\varnothing$, we have $\mathrm{int}\left(\bigcup_{n=1}^\infty A_n\right)=\varnothing$. [Recall that $\mathrm{int}(A)$ is the {\it interior} of $A$. This is the union of all open sets contained in $A$. So $\mathrm{int}(A) = \varnothing$ if $A$ contains no non-empty open sets.] In Baire's terminology, a set $A$ is of {\bf first category} if it is contained in the union of a countable collection of closed sets having empty interiors. All other sets are of {\bf second category}. So a Baire space is a space whose non-empty open sets are all of the second category.
\item[Baire Category Theorem:] compact Hausdorff spaces are Baire spaces.\\ \force\hfill Also, complete metric spaces are Baire spaces.
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