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\lhead{\large \textbf{Math 461}}
\chead{\Large \textbf{Homework 5}}
\rhead{\large \textbf{DUE 10/8/2023}}
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{\Large\underline{\textbf{Exercises:}}} (\S $14$, \S $15$, \S $16$, and \S $17$)



\begin{enumerate}

\item Let $X$ be an ordered set (with at least two elements) equipped with the order topology. For a subspace $Y\subset X$, show that the collection $\mathcal{S}$ consisting of sets of the form $Y\cap (-\infty,a)$ or $Y\cap (a,+\infty)$ for $a\in X$ form a subbasis for the subspace topology on $Y$.

\item Let $X$ and $Y$ be topological spaces. A function $f\colon X\to Y$ is called an \textbf{open map} if for every open subset $U\subset X$ one has that its image $f(U)$ is open in $Y$.  
	\begin{enumerate}
	\item Equip $X\times Y$ with the product topology. Show that the coordinate projections $\pi_1\colon X\times Y\to X$ and $\pi_2\colon X\times Y\to Y$ are open maps.
	
	\item Let $\mathcal{B}$ be a basis for the topology on $X$ and suppose $f(B)$ is open for all $B\in \mathcal{B}$. Show that $f$ is an open map.
	
	\item Show that the previous part does not hold for subbases. [\textbf{Hint:} consider the function $f\colon \mathbb{R}\to \mathbb{R}$ with $f(0)=1$ and $f(x)=|x|$ if $x\neq 0$ where $\mathbb{R}$ has the standard topology.]
	
	\end{enumerate}

\item Show that if $Y$ is a topological subspace of $X$, and $A \subset Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology that it inherits as a subspace of $X$.
	
\item Equip $\mathbb{R}$ with the standard topology.
	\begin{enumerate}
	\item Show that the subspace topology on $\{\frac1n\mid n\in\mathbb{N}\}\subset \mathbb{R}$ is the discrete topology.
	
	\item Show that the subspace topology on $\{0\}\cup\{\frac1n\mid n\in\mathbb{N}\}$ is \textbf{not} the discrete topology.
	\end{enumerate}
	
\item Let $\mathcal{C}$ be a collection of subsets of $X$. Assume that $\emptyset, X\in \mathcal{C}$ and that finite unions and arbitrary intersections of sets in $\mathcal{C}$ are in $\mathcal{C}$. Show that the collection $\mathcal{T}:=\{X\setminus C \mid C\in\mathcal{C}\}$ is a topology on $X$ and that the collection of closed sets in this topology is $\mathcal{C}$.

	
\end{enumerate}




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