Finding Errors in Proof for 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?

Homework Equations

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}}

 

[MathematicalPhysicist]

Your fourth iff is wrong.

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]

Your fourth iff is wrong.

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.

Oh yeah. I think I know what you mean. i_{0} depends on epsilon. Thanks.