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\noindent
\parbox{1.95in}{\bf Math 4710/5710}
\hfill {\Large \bf Homework \#10} \hfill
\parbox{1.95in}{\bf \hfill Due: Wed., Apr. 30${}^\mathbf{th}$\!, 2025}

\vspace{0.2in}

Recall that a {\bf net} in a topological space $X$ is a function from a {\bf directed set} $J$ (i.e. $J$ is partially ordered by $\leq$ and for every $i,j \in J$ there exists some $k \in J$ such that $i \leq k$ and $j \leq k$) to $X$. Moreover, we say that a net $(x_j)_{j \in J}$ in $X$ {\bf converges} to $x \in X$ if for every neighborhood $U$ of $x$ there exists some $N \in J$ such that $x_j \in U$ for all $j \geq N$. This is denoted: $x_j \to x$.

\vspace{0.05in}

Also, recall that a function $f:X \to Y$ (where $X$ and $Y$ are topological spaces) is said to be {\bf continuous at $x \in X$} if for every neighborhood $U$ of $f(x)$ there exists some neighborhood $V$ of $x$ such that $V \subseteq f^{-1}(U)$. 

\vspace{0.1in}

\begin{description}

\item[\#1 Continuity via Nets:] Let $X$ and $Y$ be topological spaces.
\begin{enumerate}[(a)]
\item Let $f:X \to Y$ be continuous at $x \in X$ and let $(x_j)_{j \in J}$ be a net in $X$ such that $x_j \to x$.\\ \force\hfill
Show the net $(f(x_j))_{j \in J}$ converges to $f(x)$.

\vspace{0.05in}

\item {\bf [Grad.]} Suppose that for any net $(x_j)_{j \in J}$ such that $x_j \to x$, we have $f(x_j) \to f(x)$.\\ \force\hfill 
Show $f$ is continuous at $x$.

\vspace{0.05in}

{\it Hint:} Suppose that $f$ is not continuous at $x$. Build a net which converges to $x$ but whose image does not converge to $f(x)$. You may find the proof that ``$x$ belongs to the closure of a set $A$ if and only if there is a net in $A$ converging to $x$'' helpful/instructive here.
\end{enumerate}

\vspace{0.05in}

Putting these results together, we have:\\ \force\hfill $f$ is continuous at $x$ if and only if for every net such that $x_i \to x$, we have $f(x_i) \to f(x)$. \hfill\force
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\vfill

Recall that $\mathcal{F} \subseteq \mathcal{P}(X)$ is a {\bf filter} in $X$ if $\mathcal{F} \not= \varnothing$, $\varnothing \not\in \mathcal{F}$, $A,B \in \mathcal{F}$ implies $A \cap B \in \mathcal{F}$, and $A \subseteq B$ where $A \in \mathcal{F}$ implies $B \in \mathcal{F}$. Moreover, $\mathcal{F}$ is said to be an {\bf ultrafilter} if $\mathcal{F}$ is a maximal filter (i.e. it is not properly contained in any other filter). This is equivalent to the condition that for any $A \subseteq X$ we have $A \in \mathcal{F}$ or $X-A \in \mathcal{F}$.

\vspace{0.1in}

\begin{description}
\item[\#2 Filtering out the Finite:] Let $\mathcal{F} = \{ A \subseteq X \;|\; X-A \mbox{ is finite} \}$. Show $\mathcal{F}$ is a filter if and only if $X$ is infinite.

\vspace{0.05in}

\force \hspace*{0.875in} Moreover, in the case that $\mathcal{F}$ is a filter (i.e., $X$ is infinite), explain why $\mathcal{F}$ is never an ultrafilter.
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\vspace{0.1in}

\vfill

Recall that a topological space $X$ is called {\bf completely regular} if for every closed set $A \subseteq X$ and every point $x \in X$ such that $x \not\in A$ there exists a continuous function $f:X \to [0,1]$ such that $f(x)=0$ and $f(a)=1$ for all $a \in A$ (i.e. closed sets and point can be separated by continuous functions). 

\vspace{0.1in}

\begin{description}

\item[\#3 A Completely Regular Compact Problem:] Let $X$ be a completely regular space. Let $A, B \subseteq X$ be disjoint closed sets and assume $A$ is compact. Prove there exists a continuous function $f:X \to [0,1]$ such that $f(A)=\{0\}$ and $f(B)=\{1\}$. 

\vspace{0.1in}

{\it Comment:} This falls under our philosophy that compact sets behave like points.  

\vspace{0.05in}

{\it Hint:} Since $X$ is completely regular, we can separate each point $a \in A$ from the closed set $B$ with a continuous function $f_a$. Notice that the inverse image of the open interval $(-1/2,1/2)$ under the map $f_a$ is an open set containing $a$. This should give you an open cover of $A$. Averaging some finite collection of functions should give you your desired function.   

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