% unused problems at end % LaTex2e \documentclass[12pt]{article} \usepackage{amssymb,latexsym} \pagestyle{empty} \newcount\probno \probno=0 \def\prob{\advance\probno by 1 \bigskip\par\noindent\the\probno . } \def\zz{{\mathbb Z}} \def\rr{{\mathbb R}} \def\aa{{\cal A}} \def\bb{{\cal B}} \def\cc{{\cal C}} \begin{document} \begin{flushright} A. Miller \\ M551 \\ Spring 2001\\ Final\\ \end{flushright} Open Book Open Notes Prove {\bf three} the following statements. One correct solution is worth more than three wrong ones. Please use a separate sheet of paper for each proof. Rewrite your solutions carefully inserting the reasons why what you say is true. \bigskip Don't hand in a ``solution" which you know is incorrect and say nothing about it. If part of your solution is incorrect and you know it, say so. If you just hand in a proof that you know is incorrect and you say nothing, then I will assume that you don't know what you have written is incorrect. \bigskip \prob Suppose D is dense in X and U is open in $X$. Prove that $U \subseteq \overline{U\cap D}$ \prob A point $x\in X$ is a clustre point of $A\subseteq X$ iff every open neighborhood of $x$ contains uncountably many points of $A$. Suppose $X$ is a Lindel\"of space and A is an uncountable closed subspace of $X$. Prove that at least one point of $A$, ($x\in A$) is a clustre point of $A$. \prob Suppose $\tau$ and $\tau^\prime$ are topologies on the same set $X$. Suppose $(X,\tau)$ is a Hausdorff topology, $\tau\subseteq \tau^\prime$, and $(X,\tau^\prime)$ is a compact space. Prove that $\tau=\tau^\prime$. \prob Suppose $C\subseteq X$ is a connected subspace of the connected space $X$. Suppose $X-C=A\cup B$ where $\overline{A}\cap B={A}\cap \overline{B}=\emptyset$. Prove that both $C\cup A$ and $C\cup B$ are connected. \prob Suppose $X$ is regular and $Q=\{x_n:n<\omega\}$ is a countable discrete subset of $X$, i.e., $Q$ is discrete in the subspace topology. Prove there exist open sets $\{U_n :n<\omega\}$ which are pairwise disjoint and separate $Q$, i.e., $x_n \in U_n$ for each $n$ and $U_n\cap U_m=\emptyset$ for each $n\not=m$. \prob Suppose $X\subseteq {\mathbb R}_l$ (the Sorgenfrey line) and $X$ is compact in the subspace topology. Prove that $X$ is countable. \newpage 1. Let $V=U -\overline{U\cap D}$. The set $V$ is open, hence if it is nonempty it must meet the dense set $D$. But this is a contradiction, since $V\subseteq U$ and so $V\cap D\subseteq U\cap D$, but $V$ is disjoint from $U\cap D$. 2. Closed subspaces of Lindelof spaces are Lindelof. If there is no clustre point in $A$ then $A$ can be covered by countably many open sets $U$ with the property that $U\cap A$ is countable. But then $A$ would be the countable union of countable sets, hence countable. 3. If $C$ is $\tau^\prime$-closed, then it is compact. If it is $\tau^\prime$-compact it is $\tau$-compact, since $\tau\subseteq \tau^\prime$. In a Hausdorff space compact sets are closed, hence $C$ is $\tau$-closed. Since $\tau^\prime$-closed implies $\tau$-closed, by taking compliments $\tau^\prime\subseteq \tau$. 4. Suppose $C\cup A$ is not connected. Then there exists nonempty $E_1,E_2$ such that $C\cup A=E_1\cup E_2$ and $E_1\cap \overline{E_2}= \overline{E_1}\cap E_2=\emptyset$. Since $C$ is connected it cannot be separated by the $E_i$, so assume $E_1=C\cup A_1$ and $E_2=A_2$ where $A_1\cup A_2=A$. Then $A_3=C\cup A_1\cup B$ and $A_2$ separate $X$, since $A_2\cup A_3=X$ and $A_3\cap \overline{A_2}= \overline{A_3}\cap A_2=\emptyset$. 5. Inductively construct $U_n$ so that $\overline{U_n}$ are pairwise disjoint, $x_n\in U_n$, and $x_m\notin \overline{U_n}$ for all $m>n$. Suppose we already have $U_0,U_1,\ldots,U_n$. Since the set $Q$ is discrete there exists $W$ an open neighborhood of $x_{n+1}$ which does not contain any $x_m$ for $m\not=n+1$. Since $x_{n+1}$ is not in the closed set $\bigcup_{m\leq n}\overline{U_m}$, we can shrink $W$ if neccessary so that it is disjoint from it. Since the space is regular we can find an open set $U_{n+1}$ with $$x_{n+1}\in U_{n+1} \subseteq \overline{U_{n+1}}\subseteq W$$ 6. First note that if $X$ is an uncountable set of reals it must contain a strictly increasing sequence $a_n$ for $n\in\omega$. Letting $a$ be the sup of this sequence we see that $X$ is covered by $\{(-\infty,a_n):n\in\omega\}\cup \{[a,\infty)\}$ which has no finite subcover. \par To see that an uncountable $X$ must contain a strictly increasing sequence, suppose not. Then for every real $a$ there exists $b