Problem 30 in the Exercises following Chapter 2 in Baby Rudin: How to immitate the proof of Theorem 2.43?

Question

Here's problem 30, the very last one, in the Exercises after Chapter 2 in Walter Rudin's Principles of Mathematical Analysis, 3rd edition:

Imitate the proof of Theorem 2.43 to obtain the following result:

If $\mathbb{R}^k = \cup_{n=1}^\infty F_n$, where each $F_n$ is a closed subset of $\mathbb{R}^k$, then at least one $F_n$ has a non-empty interior.

Equivalent statement:

If $G_n$ is a dense open subset of $\mathbb{R}^k$, for each $n = 1, 2, 3, \ldots$, then $\cap_{n=1}^\infty G_n$ is not empty (in fact, it is dense in $\mathbb{R}^k$).

(This is a special case of Baire's Theorem; ...)

Now I have seen a proof of this result in a solutions manual to Babay Rudin, but I wonder how to obtain a proof like that of Theorem 2.43, as required by Rudin?

I think I have fully comprehended the proof of Theorem 2.43 and can even reproduce it with some effort.

However, I'm unable to make a connection between that proof and the proof I've seen of the above result.

Answer 1

Soln: Let U be any nonempty open set in $\mathbb R^k$. Set $U_0=U$.

Since $G_1$ is open and dense, $U_0\cap G_1$ is a $\textbf{nonempty}$ open set.

Let $x\in U_0 \cap G_1$ and $r>0$, such that $\overline{B(x,r)} \subsetneq U_0\cap G_1$. Set $U_1 = B(x,r)$ and $K_1=\overline{U_1}$. Then $K_1$ is compact, as bounded closed sets are compact in $\mathbb R^n$.

We inductively construct $U_n,K_n$ as follows.

Suppose $U_m,K_m$ has been constructed, and $U_m,K_m$ are nonempty. Then $U_m\cap G_{m+1}$ is a nonempty open set as $G_{m+1}$ is open and dense. Let us choose a $x\in U_m\cap G_{m+1}$ and $r>0$, such that $\overline{B(x,r)} \subsetneq U_m\cap G_{m+1}$. Then set $U_{m+1} = B(x,r)$ and $K_n = \overline{U_{m+1}}$.

Then we have a decreasing sequence of $\textbf{nonempty}$ open sets $$\cdots\subsetneq U_{n+1}\subsetneq U_n \subsetneq \cdots U_1\subsetneq U_0=U$$

And hence we have a decreasing sequence of $\textbf{nonempty}$ compact sets $$\cdots K_{n+1}\subsetneq K_n \subsetneq \cdots K_1\subsetneq U_0=U$$ as $K_i$ are closed, $K_i\subset K_1$ and $K_1$ is compact. We also have $U_n\subsetneq K_n \subsetneq\cap_{i=1}^nG_n$ and $K_n$ is nonempty as $U_n$ is nonempty.

A family of compact sets has $\textbf{finite intersection property(FIP)}$. In particular the family $\{K_n\}$ has FIP. As $\cap_{i=1}^nK_i = K_n\neq \emptyset$, we have $\cap_{i=1}^\infty K_n$ is $\textbf{noempty}$. Let $x\in \cap_{i=1}^\infty K_n$. Then $x\in K_{n}$ for all $n$, so $x\in G_n$ for all $n$. Also $x\in U_0=U$. So $x\in U\cap(\cap_{i=1}^\infty G_n)$.

Hence $U\cap(\cap_{i=1}^\infty G_n)$ is nonempty. (It may not be open)

As $U$ was an arbitrary non empty open set, we can conclude that $\cap_{i=1}^\infty G_n$, intersects an arbitrary nonempty open set in a nonempty set. Hence $\cap_{i=1}^\infty G_n$ is a dense subset of $\mathbb R^k$. In particular it is nonempty. END

Comments:

0a. A subset of a topological space is dense iff it intersects every nonempty open set in a nonempty set.

0b. A set has empty interior if and only if its complement is dense.

1. The only place we used the fact that the ambient metric space is $\mathbb R^k$ is to use that $\overline{B(x,r)}$ is compact. So this statement must hold good for metric spaces where $\overline{B(x,r)}$ for finite $r$ is compact. Note this last condition is equivalent to the condition that bounded and closed sets are compact.

2. The statement in the question is equivalent to : if $\{F_n\}$ is a family of closed sets in $\mathbb R^k$ each of which has empty interior then $\cup_{i=1}^\infty F_n$ has empty interior. In particular it is not equal to the ambient metric space. This is an important fact. A corollary of this is: $\mathbb R^n$ is not the the union of countably many hypersurfaces. Another corollary is: a manifold is not a countable union of its lower dimensional submanifolds. This follows since a manifold is locally homeomorphic to $\mathbb R^n$, we can apply this fact locally.

3. A subset of a metric space is nowhere dense iff the closure of this subset has empty interior.

4. A metric space which is the union of countably many nowhere dense sets is said to be of the first category.

5. A metric space which is not of the first category is said to be of the second category.

6. The Baire category theorem says that complete metric spaces are of the second category. $\mathbb R^k$ is a complete metric space, so this theorem says that $\mathbb R^k$ is not a countable union of closed sets with empty interiors by Comment 3. But the proof of this theorem is more involved.