Thread: Finding an error in a proof (topology)

1. Finding an error in a proof (topology)

I am trying to find an error in the below proof. Any advice will be highly appreciated.

Proof: $\overline{\cup A_{\alpha}} \subset \cup\bar{A_{\alpha}}$

If $\{A_{\alpha}\}$ is a collection of sets in X and if $x \in \overline{\cup A_{\alpha}}$, then every neighborhood $U$ of x intersects $\cup A_{\alpha}$. Then $U$ must intersect some $A_{\alpha}$, so that x must belong to the closure of some $A_{\alpha}$. Therefore, $x \in \cup\bar{A_{\alpha}}$.

2. Originally Posted by aliceinwonderland
I am trying to find an error in the below proof. Any advice will be highly appreciated.

Proof: $\overline{\cup A_{\alpha}} \subset \cup\bar{A_{\alpha}}$

If $\{A_{\alpha}\}$ is a collection of sets in X and if $x \in \overline{\cup A_{\alpha}}$, then every neighborhood $U$ of x intersects $\cup A_{\alpha}$. Then $U$ must intersect some $A_{\alpha}$, so that x must belong to the closure of some $A_{\alpha}$. Therefore, $x \in \cup\bar{A_{\alpha}}$.
It's true that every neighbourhood U of x must intersect some $A_{\alpha}$. However, different neighbourhoods may intersect different $A_{\alpha}$s. Your argument assumes that each U intersects a single $A_{\alpha}$.

Here's an example. For n=1,2,3,..., let $A_n=(1/n,1]$. Then $\cup A_n = (0,1]$, and $\overline{\cup A_n} = [0,1]$. But $\bar{A}_n=[1/n,1]$, and $\cup \bar{A}_n = (0,1]$. So $0\in\overline{\cup A_n}$ but $0\notin\cup \bar{A}_n$.