Question

$\underset{1}{\overset{\infty }{\cup }}{E}_{i}F=\left(\underset{1}{\overset{\infty }{\cup }}{E}_i\right)F$and$\underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)=\left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F$

Short answer

$$\begin{gathered} \underset{1}{\overset{\infty }{\cup }}{E}_{i}F=\left(\underset{1}{\overset{\infty }{\cup }}{E}_i\right)F \\ \underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)=\left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F \end{gathered}$$

Step-by-step solution

Given Information.

$\underset{1}{\overset{\infty }{\cup }}{E}_{i}F=\left(\underset{1}{\overset{\infty }{\cup }}{E}_i\right)F$and$\underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)=\left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F$

Explanation.

first part

let$x\in \left(\underset{1}{\overset{\infty }{\cup }}{E}_i\right)F$. Then$x\in F$and$x\in \underset{1}{\overset{\infty }{\cup }}{E}_{i}x\in \underset{1}{\overset{\infty }{\cup }}{E}_i$implies,$x\in E_i$for some$i=j$. This implies that$x\in E_{j}F$. Thus $x\in E_{i}F$for some$i$and hence$x\in \underset{1}{\overset{\infty }{\cup }}{E}_{i}F$. Therefore,$\left(\underset{1}{\overset{\infty }{\cup }}{E}_i\right)F\subseteq \underset{1}{\overset{\infty }{\cup }}{E}_{i}F$

conversely,

let$x\in \underset{1}{\overset{\infty }{\cup }}{E}_{i}F$. Then $x\in E_{i}F$for some$i=j$. This implies that $x\in F$and$x\in E_j$.

i.e. $x\in F$and$x\in E_i$for somewidth="6">i. i.e.$x\in F$and$x\in \underset{1}{\overset{\infty }{\cup }}{E}_i$.

Hence, localid="1649063117247" $x\in$localid="1649063128981" $\left(\underset{1}{\overset{\infty }{\cup }}{E}_i\right)F$

Therefore, from the above two arguments,$\underset{1}{\overset{\infty }{\cup }}{E}_{i}F=\left(\underset{1}{\overset{\infty }{\cup }}{E}_i\right)F$

Explanation.

second part

let$x\in \left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F$. This means $x\in F$or$x\in \underset{1}{\overset{\infty }{\cap }}{E}_i$. If $x\in F$then $x\in F\cup E_i$for all $i$and hence$x\in \underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)$. If $x\in E_i$for all$i$then again, $x\in F\cup E_i$for all$i$and hence$x\in \underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)$. Therefore$\left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F\subseteq \underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)$.

conversely,

say$x\in \underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)$. Then $x\in F\cup E_i$for all$i$. If$x\in F$then clearly$x\in \left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F$. If not, then since $x\in F\cup E_i$all$i$, $x$has to be in $E_i$for all$i$. i.e.$x\in E_i$for all $i$and hence$x\in \underset{1}{\overset{\infty }{\cap }}{E}_i$. Therefore, $x\in \left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F$and$\underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)\subseteq \left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F$.

from the above two arguments,$\underset{1}{\overset{\infty }{\cap }}\left(E_i\cup F\right)=\left(\underset{1}{\overset{\infty }{\cap }}{E}_i\right)\cup F$