# Rudin Problem 2.7

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1. Oct 2, 2014

### Mogarrr

1. The problem statement, all variables and given/known data

Let $A_1, A_2, A_2,...$ be subsets of a metric space.
(a) If $B_{n}=\bigcup_{i=1}^{n}A_{i}$, prove that $\bar{B_{n}}= \bigcup_{i=1}^{n}\bar{A_{i}}$ for n=1,2,3,...
(b) If $B=\bigcup_{i=1}^{\infty}A_{i}$, prove that $\bar{B} \subset \bigcup_{i=1}^{\infty}\bar{A_{i}}$,
Show by example this inclusion is proper.

2. Relevant equations

If E is a subset of a metric space, then the closure of E,
$\bar{E}=E \bigcup \acute{E}$, where $\acute{E}$ is the set of all limit points of E.

For a subset E of a metric space X, a point $p \in X$ is a limit point if and only if
$\forall \epsilon > 0, \exists q \in B(p;\epsilon) \bigcap E : q \neq p$

3. The attempt at a solution

Now I think I've got part (a) and (b) nailed down, but I'm having trouble thinking of an example for a proper subset. Here's what I have thus far...

Suppose $x \in \bar{B_{n}}$, then $x \in B_{n}$ or x is a limit point for $B_{n}$.
If $x \in {B_{n}}$, then $x \in \bigcup_{i=1}^{n} A_{i}$, then $\exists i, x \in A_{i}$, so $\exists i, x \in \bar{A_{i}}$.
If x is a limit point, then
$\forall \epsilon > 0, \exists y \in B(x;\epsilon) \bigcap B_{n}: y \neq x$, so $\forall \epsilon > 0, \exists y \in B(x;\epsilon) \bigcap (\bigcup_{i=1}^{n} A_{i}): y \neq x$, then
$\exists i,\forall \epsilon > 0, \exists y \in B(x;\epsilon) \bigcap A_{i}: y \neq x$, then
$\exists i, x\in \bar{A_{i}}$, so $x \in \bigcup_{i=1}^{n}\bar{A_{i}}$

Suppose $x \in \bigcup_{i=1}^{n}\bar{A_{i}}$, then $\exists i, x \in A_{j}$ or for some i x is a limit point.
If $\exists i, x \in A_{i}$, then $x \in \bigcup_{i=1}^{n}A_{i}=B_{n}$.
If x is a limit point for some A sub i, then x is a limit point for $\bigcup A_{i}=B_{n}$.

For part (b), I did the same thing, but replaced $B_{n}$ with $B$ and in the union I replaced n with infinity.

Now, I'm having some difficulty coming up with an example that would show B is a proper subset of the union of an infinite number of A sub i's.

Any hints or corrections would be much appreciated.

2. Oct 2, 2014

### Ray Vickson

I can see a fixable flaw in your argument for (a). For simplicity, take $n = 2$; that is not a loss of generality, since you can then get the general case of finite $n$ by induction.

Anyway, say $x \in \bar{B}_2$ is in the boundary of $B_2$ but not in $B_2$ itself. Let $\epsilon_1 > \epsilon_2 > \cdots \to 0$. For $\epsilon_i$ there is a point $x_i \in B_2 \cap N(x,\epsilon_i)$. It is conceivable that we could have, for example, $x_1 \in \ A_1, x_2 \in A_2, x_3 \in A_1, x_4 \in A_2, \ldots$ so the sequence $\{x_i\}$ is not in a single set $A_1$ or $A_2$, but, instead, oscillates back and forth between the two. In that case you would need to do more work to conclude that $x$ is in $\bar{A}_1 \cup \bar{A}_2$. It can be done, but some extra steps are needed; I think the fact that you are working an a metric space matters a lot here.

3. Oct 2, 2014

### pasmith

Are you sure the inclusion in part (b) is the right way round? Certainly one can show by example that $$\bigcup_{i=1}^\infty \bar{A}_i \subsetneq \bar{B}$$ by taking a bijection $f : \{1, 2, 3, \dots \} \to \mathbb{Q}$ (such a bijection exists because the rationals are countable) and setting $A_i = \{f(i)\} \subset \mathbb{R}$.

4. Oct 2, 2014

### Mogarrr

So what you're saying is that just because $\forall \epsilon >0,\exists y \in N(x;\epsilon) \bigcap \bigcup_{i=1}^{n} A_{i}$, I can't conclude that $\exists i, \forall \epsilon >0,\exists y \in N(x;\epsilon) \bigcap A_{i}$?

Would I have to say that for a sequence of epsilons, there is a y in the intersection of the Neighborhood with center x and radius epsilon? I thought about distributing the intersection/union, but that doesn't make much sense to me, intuitively.

5. Oct 2, 2014

### Mogarrr

pasmith,

Wow, really nice example. Am I correct in saying "because each of the singletons have no limit points, the union of the A sub i's are the rationals, but the closure of B is the reals"?

And yes, I was wrong. It should be....

Show by an example, that this inclusion $B \supset \bigcup_{i=1}^{\infty} A_{i}$ can be proper.

6. Oct 2, 2014

### pasmith

The point is not that a singleton has no limit points, but that it contains its only limit point and is therefore closed, so that $\bigcup A_i = \bigcup \bar{A}_i$. Otherwise you are correct, although you need to explain briefly why $\bigcup_{i=1}^\infty \{f(i)\} = \mathbb{Q}$.

Last edited: Oct 2, 2014
7. Oct 2, 2014

### Ray Vickson

You probably can conclude it, but it does not follow right away from what you did; you need some extra steps.