Got it now after plenty of sleep. 1 $\displaystyle \in S$ and {1} $\displaystyle \in S$ imply {1}$\displaystyle \subset S$

Manx, you are a good friend because you put up with me.

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technically saying {1} E S is wrong. 1 E S given S {

1,{2}, {1,2}}.

{1} isnt an element of S. but i think i know where your confusion comes from. a subset gets

{} regardless so

{1

} represents the 1 in S.

{{2}

} represents the {2} in S and

{1,{2}

} represents the 1,{2} in S. on a test if you say {1} E S that would be wrong because there is no element {1} in S there IS an element of 1 in S. Saying that there is a

{1} E S means that this would be S {

{1},{2},{1,2}} saying 1 E S is saying there is a 1 in S which there is {

1,{2},{1,2}}