# 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: $\displaystyle \overline{\cup A_{\alpha}} \subset \cup\bar{A_{\alpha}}$

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

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

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

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

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