# Baby Rudin Problem 2.7

julypraise

## Homework Statement

I've proved that if $B = \bigcup_{i=1}^{\infty} A_{i}$ then $\overline{B} = \bigcup_{i=1}^{\infty} \overline{A_{i}}$ but it should not be right. So could you find errors on my reasoning?

## The Attempt at a Solution

Observe $$x \in \overline{B}$$

iff for every $$\epsilon>0 \quad B(x;\epsilon) \cap B \neq \emptyset$$

iff $$B(x;\epsilon) \cap \bigcup_{i=1}^{\infty} A_{i} \neq \emptyset$$

iff $$B(x;\epsilon) \cap A_{i_{0}} \neq \emptyset$$ for some $i_{0} \in \mathbb{Z}^{+}$

iff $$x \in \overline{A_{i_{0}}}$$

iff $$x \in \bigcup_{i=1}^{\infty} \overline{A_{i}}$$

Gold Member

it should be if $x \in \bar{A_i_0}$ then $B(x,\epsilon)\cap A_{i_0}$.

I mean if $B(x,\epsilon) \cap A_{i_0}$ \forall \epsilon >0, x can still be outside of A_{i_0}.

Take another set such A_j such that x is in it but not in A_i_0 but they intersect each other, and such that whatever nbhd of x we pick it intersects A_i_0.

julypraise
it should be if $x \in \bar{A_i_0}$ then $B(x,\epsilon)\cap A_{i_0}$.
I mean if $B(x,\epsilon) \cap A_{i_0}$ \forall \epsilon >0, x can still be outside of A_{i_0}.