2

I am not sure if the way I tried proving this problem is correct and so I would appreciate the feedback.

My proof:

(i) Let x $\in B \cap\bigcup_{\alpha\in\Delta}A_\alpha $. Then x $\in B$ and $x\in \bigcup_{\alpha\in\Delta}A_\alpha$. Then X $\in B$ and x $\in A_\alpha$ for some $\alpha\in\Delta$. Thus, x ($\in$ B $\cap$ $A_\alpha$) for some $\alpha \in \Delta$. Thus x $\in\bigcup_{\alpha\in\Delta}(A_\alpha\cap B)$. Therefore,$B\cap\bigcup_{\alpha\in\Delta}A_\alpha \subseteq\bigcup_{\alpha\in\Delta}(A_\alpha\cap B)$

(ii) Let X $\in \bigcup_{\alpha\in\Delta}(A_\alpha\cap B)$. Then x $\in (B \cap A_\alpha)$ for some $ \alpha\in\Delta$. Then x $\in B$ and $x \in A_\alpha$ for some $\alpha \in \Delta$. Thus, x $\in B \cap\bigcup_{\alpha\in\Delta}A_\alpha$. Therefore, $ \bigcup_{\alpha\in\Delta}(A_\alpha\cap B) \subseteq B \cap \bigcup_{\alpha\in\Delta}A_\alpha $.

Thus from (i) and (ii), B $\cap$ $\bigcup_{\alpha\in\Delta}A_\alpha = \bigcup_{\alpha\in\Delta}(A_\alpha\cap B)$.

Edited cuz of typo With second sentence of (ii).

Jr194
  • 109
  • 2
    Looks good to me! – angryavian Nov 12 '19 at 02:55
  • 1
    This looks fine. But there is a mistake in the second part. You write "Let $x\in \bigcup_{\alpha\in\Delta}(A_\alpha\cap B)$. Then $x\in (B \cup A_\alpha)$ for some $\alpha\in\Delta$" instead of "Let $x\in \bigcup_{\alpha\in\Delta}(A_\alpha\cap B)$. Then $x\in (B \color{red}{\cap} A_\alpha)$ for some $\alpha\in\Delta$" – azif00 Nov 12 '19 at 03:36
  • Tnx for noticing, that was a typo on my part, and I fixed it just now. – Jr194 Nov 12 '19 at 03:51

0 Answers0