Proving No Set Contains All Sets Without Russell's Paradox

[nikkor180]

Greetings: I am attempting to prove that no set contains all sets without Russell's paradox. What I have thus far is this:

Let S be an arbitrary set and suppose S contains S. If X is in S for some X not=S, then S - S cannot be empty. But this is a contradiction; hence if S contains S, then S contains only S. Thus S cannot contain all sets.

Is this argument valid?

Thank you.

Rich B. (note: I am not a student; this is not homework)

 

[Evgeny.Makarov]

Could you write your reasoning with formulas?

Let S be an arbitrary set and suppose S contains S.

Do you mean $S\in S$ or $S\subseteq S$?

If X is in S for some X not=S, then S - S cannot be empty.

Why?

 

[nikkor180]

Evgeny.Makarov: Thank you for responding to my post.

1) I mean "S is an element of S" (not subset).

Regarding "If X is in S for some X not=S, then S - S cannot be empty", I need to ponder that further. I will respond soon.

Thanks again,

Rich B.

BTW, can you tell me how to access symbols. I clicked on "element of" in the quick index but such came through in preview as \in.

 

[MarkFL]

...
BTW, can you tell me how to access symbols. I clicked on "element of" in the quick index but such came through in preview as \in.

Hello, Rich! (Wave)

To get $\LaTeX$ symbols/commands to render as such, they need to be enclosed with tags. The simplest way to do this is to click the \(\displaystyle \large\Sigma\) button on the toolbar, which will generate [MATH][/MATH] tags for you, with the cursor located between the tags, ready for your input of code. :)

 

[Evgeny.Makarov]

BTW, can you tell me how to access symbols.

You should enclose formulas in dollar signs or [MATH]...[/MATH] tags. The tags can be inserted either manually or by clicking the $\Sigma$ button on the toolbar above the edit region. You can also click "Reply with Quote" to see how other users typed their messages. For more on formulas see http://mathhelpboards.com/latex-tips-tutorials-56/.

 

[nikkor180]

Thanks folks.

 

[Evobeus]

Greetings: I am attempting to prove that no set contains all sets without Russell's paradox. What I have thus far is this:

Let S be an arbitrary set and suppose S contains S. If X is in S for some X not=S, then S - S cannot be empty. But this is a contradiction; hence if S contains S, then S contains only S. Thus S cannot contain all sets.

Is this argument valid?

Thank you.

Rich B. (note: I am not a student; this is not homework)

This is clearly evident from Cantor's Diagonal Theorem which leads to the conclusion that there is no greatest cardinal number ad hence no set of all sets.
For an easy routine work consider power set.