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\noindent
\parbox{1.95in}{\bf Math 4710/5710}
\hfill {\Large \bf Homework \#4} \hfill
\parbox{1.95in}{\bf \hfill Due: Mon., Feb. 17${}^\mathbf{th}$\!, 2025}

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%\noindent {\bf\large Name:} \quad \underline{\large\qquad This is my Name \qquad}

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\noindent
Please remember when submitting any work via email or in person to\dots\\
\force \hfill \fbox{\fbox{\bf\large\color{red}PUT YOUR NAME ON YOUR WORK!}} \hfill \force

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\begin{description}
\item[\#1 Based Question:] We have the following lemma (Munkres' Lemma 13.2 on page 80):

\LEMMA Let $X$ be a topological space (with topology $\mathcal{T}$). Suppose $\mathcal{C} \subseteq \mathcal{T}$ (i.e., $\mathcal{C}$ is a collection of open sets) such that for each open set $U$ (i.e., $U \in \mathcal{T}$) and $x \in U$, there is some $C \in \mathcal{C}$ such that $x \in C \subseteq U$ (i.e., elements of $\mathcal{T}$ are unions of elements of $\mathcal{C}$). Then $\mathcal{C}$ is a basis that generates $\mathcal{T}$.

\begin{enumerate}[(a)]
   \item Use the above lemma to show $\mathcal{B} = \{(a,b) \;|\; a<b \text{ where } a,b \in \mathbb{Q}\}$ is a basis for $\mathbb{R}$\\ \force\hfill (i.e., the reals equipped with the standard topology).
   
   \item We can also use $\mathcal{B}_{\mathrm{std.}} = \{ (a,b) \;|\; a<b \text{ where } a,b\in \mathbb{R} \}$ as a basis for $\mathbb{R}$. 
   
            Compare the sizes of $\mathcal{B}$ and $\mathcal{B}_{\mathrm{std.}}$. Show that $\mathbb{R}$ cannot have a basis smaller than $\mathcal{B}$.
            
   \item Show $\mathcal{C} = \{ [a,b) \;|\; a<b \text{ where } a,b \in \mathbb{Q}\}$ is a basis for the set of real numbers. 
   
   \item {\bf [Grad.]} Show $\mathcal{C}$ does not generate the lower limit topology on $\mathbb{R}$.
\end{enumerate}             

\item[\#2 Real Weirdness:] We discussed the lower limit topology on the reals: $\mathbb{R}_\ell = \mathbb{R}$ has its topology generated by $\mathcal{B}_\ell = \{ [a,b) \;|\; a<b \text{ where } a,b \in \mathbb{R}\}$. Similarly, one can define the upper limit topology on the reals: $\mathbb{R}_u=\mathbb{R}$ where its topology is generated by $\mathcal{B}_u = \{ (a,b] \;|\; a<b \text{ where } a,b \in \mathbb{R}\}$.

\begin{enumerate}[(a)]
\item In class, we showed $\mathbb{R}_\ell$'s topology is strictly finer than the standard topology. Adapt this argument to show the same is true for $\mathbb{R}_u$ (i.e., $\mathbb{R}_u$'s topology is strictly finer than the standard topology). Then show the lower limit and upper limit topologies are incomparable.

\item The union of two subbases is still a subbasis. In fact, if $\mathcal{S}$ is a subbasis for $X$ and $\mathcal{S} \subseteq \mathcal{S}' \subseteq \mathcal{P}(X)$, then $\mathcal{S}'$ is still a subbasis (since $\bigcup \mathcal{S}' \supseteq \bigcup \mathcal{S}=X$ so $\bigcup \mathcal{S}'=X$). 

 However, this is not true of bases. Explain why $\mathcal{B}_\ell \cup \mathcal{B}_u$ is not a basis.

\item {\bf [Grad.]} Let $\mathcal{T}$ be a topology on $\mathbb{R}$ generated by the {\it subbasis} $\mathcal{B}_\ell \cup \mathcal{B}_u$. This will necessarily be strictly finer than both $\mathbb{R}_\ell$'s and $\mathbb{R}_u$'s topologies. (Why?) Describe $\mathcal{T}$. Better yet, what is $\mathcal{T}$?
\end{enumerate}

% Could assign intersection of topologies is a topology. Would have been a good ``from the definition'' question.


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