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


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}.

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