1. This is a very basic set theory, and I can't even get it---it so sad.

Dexel,
Some by stander urged to me cross a street, and the other stop me before I got to the other side of it. Now, I am in the middle of it waiting for a disaster to happen.

Originally Posted by Drexel28
I don't think what you are saying is correct. If $\displaystyle a\in S$ THEN $\displaystyle \{a\}\subseteq S$. You seem to be having an awful lot of confusion. Have you tried looking at what subset means abstractly? $\displaystyle E\subseteq S \text{ iff }x\in E\implies x\in S$.
To me $\displaystyle (E\subseteq S )\Longleftrightarrow(x\in E\implies x\in S)$ means

$\displaystyle (E=\{1\}$ and $\displaystyle S=\{1,\{2\}, \{1,\{2\}\}\}) \Longleftrightarrow(E\subseteq S )$

or

$\displaystyle (E=\{1\}$ and $\displaystyle S=\{1,2,3,4\}) \Longleftrightarrow(E\subseteq S )$

2. Originally Posted by Drexel28
I don't think what you are saying is correct. If $\displaystyle a\in S$ THEN $\displaystyle \{a\}\subseteq S$. You seem to be having an awful lot of confusion. Have you tried looking at what subset means abstractly? $\displaystyle E\subseteq S \text{ iff }x\in E\implies x\in S$.
I see what you are pointing at. I found where I made mistake and marked it in red:

$\displaystyle (1\in \{1\}$ and {1}$\displaystyle \in S) \Rightarrow \{1\} \subset S$ was definitely sloppy.

I meant to say $\displaystyle (1\in \{1\}$ and $\displaystyle 1 \in S) \Rightarrow \{1\} \subset S$

Thank you for watching over me.

3. [quote=canyiah;446827]
Originally Posted by novice
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.[/quote

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}}
I see you are as confused as I was at the begining before I posted the question. This thread has been longer than I expected. I have just wriggled myself out exhausted. If you go back to where I began, you will see how I got out of it.

Happy learning!

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