## 414/514 – Some topology

This is homework 3, due Monday, October 17, at the beginning of lecture.

• Recall the Baire category theorem: The intersection of countably many dense open sets in a complete metric space is dense. Recall that a $G_\delta$ set is one that can be written as the intersection of countably many open sets.

Show that in a complete metric space, the intersection of countably many dense $G_\delta$ sets is again a dense $G_\delta$ set. Is the union of two $G_\delta$ sets again a $G_\delta$ set?

Show that ${\mathbb Q}$ is not a $G_\delta$ subset of ${\mathbb R}$.

Since ${\mathbb Q}$ is countable, we can enumerate it as $a_1,a_2,\dots$. For $j\ge 1$ let $V_j=\bigcup_i(a_i-\frac1{2^{j+i}},a_i+\frac1{2^{j+i}}).$

Note that each $V_j$ is open, that ${\mathbb Q}\subseteq V_j$ for each $j$, that $V_1\supseteq V_2\supseteq\dots$, and that (for each $j$) the sum of the lengths of the intervals that make up $V_j$ is at most $2^{1-j}$.

Does this mean that ${\mathbb Q}=\bigcap_j V_j$? If yes, please provide a proof. If not, describe as concretely as possible an irrational number that belongs to this intersection.

• Recall that a nowhere dense set is a set $A$ whose closure has empty interior: $(\bar A)^\circ=\emptyset$. A set is meager (or of the first category) iff it is the union of countably many nowhere dense sets. Given sets $A$ and $B$ in ${\mathbb R}$, we say that $A=^*B$ iff their symmetric difference $(A\setminus B)\cup(B\setminus A)$ is meager. For example, the Cantor set is nowhere dense.

Show that $=^*$ is an equivalence relation.

Show that if $A$ is meager, then any subset of $A$ is also meager. Show that the union of countably many meager sets is again meager.

Show that the Baire category theorem implies that any nonempty open subset of ${\mathbb R}$ is non-meager.

Show that if $A$ is open, or closed, or a $G_\delta$ set, or a $F_\sigma$ set (a countable union of closed sets), then there is an open set $U$ such that $A=^*U$.

• The Cantor-Bendixson derivative of a closed set $X$ is defined by $X'=\{x\in X\mid x\mbox{ is a limit point of }X\}.$

Since $X$ is closed, $X'\subseteq X$. We can iterate this operation, and form $X''$, $X^{(3)}=(X'')',\dots,X^{(n+1)}=(X^{(n)})',\dots$ Note we have $X\supseteq X'\supseteq X''\supseteq\dots$

We can go even further, by letting $X^{(\infty)}=\bigcap_n X^{(n)}$ and then continuing, by setting $X^{(\infty+n+1)}=(X^{(\infty+n)})'$, etc.

Give examples of closed subsets $X$ of ${\mathbb R}$ such that $X\ne\emptyset$ but $X'=\emptyset$, or $X'\ne\emptyset$ but $X''=\emptyset$, or $\dots$ or $X^{(n)}\ne\emptyset$ but $X^{(n+1)}=\emptyset$. Can $X^{(\infty)}$ be the first empty “derivative”? How about $X^{(\infty+3)}$?

• Check that if $K_1\subseteq X_1$ and $K_2\subseteq X_2$ are compact, then so is $K_1\times K_2$ as a subset of $X_1\times X_2$.
• Show that if $X$ is a subset of ${\mathbb R}$ that is both open and closed, and $X\ne\emptyset$, then it must be the case that $X={\mathbb R}$.