91Ó°ÊÓ

$\underset{1}{\overset{∞}{∪}}{E}_{i}F=\left(\underset{1}{\overset{∞}{∪}}{E}_i\right)F$and$\underset{1}{\overset{∞}{âˆ\copyright }}\left(E_i∪F\right)=\left(\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i\right)∪F$

first part

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

conversely,

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

i.e. $x∈F$and$x∈E_i$for somewidth="6">i. i.e.$x∈F$and$x∈\underset{1}{\overset{∞}{∪}}{E}_i$.

Hence, localid="1649063117247" $x∈$localid="1649063128981" $\left(\underset{1}{\overset{∞}{∪}}{E}_i\right)F$

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

second part

let$x∈\left(\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i\right)∪F$. This means $x∈F$or$x∈\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i$. If $x∈F$then $x∈F∪E_i$for all $i$and hence$x∈\underset{1}{\overset{∞}{âˆ\copyright }}\left(E_i∪F\right)$. If $x∈E_i$for all$i$then again, $x∈F∪E_i$for all$i$and hence$x∈\underset{1}{\overset{∞}{âˆ\copyright }}\left(E_i∪F\right)$. Therefore$\left(\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i\right)∪F⊆\underset{1}{\overset{∞}{âˆ\copyright }}\left(E_i∪F\right)$.

conversely,

say$x∈\underset{1}{\overset{∞}{âˆ\copyright }}\left(E_i∪F\right)$. Then $x∈F∪E_i$for all$i$. If$x∈F$then clearly$x∈\left(\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i\right)∪F$. If not, then since $x∈F∪E_i$all$i$, $x$has to be in $E_i$for all$i$. i.e.$x∈E_i$for all $i$and hence$x∈\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i$. Therefore, $x∈\left(\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i\right)∪F$and$\underset{1}{\overset{∞}{âˆ\copyright }}\left(E_i∪F\right)⊆\left(\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i\right)∪F$.

from the above two arguments,$\underset{1}{\overset{∞}{âˆ\copyright }}\left(E_i∪F\right)=\left(\underset{1}{\overset{∞}{âˆ\copyright }}{E}_i\right)∪F$