Is the Set M a Version of Russell's Paradox?

[Unit]

Is this set a variation of Russell's paradox?

$$M = \{x : x \notin M \}$$

I understand this formulation a lot more than

$$R = \{S : S \notin S\}$$

because I don't understand how, for example, 1 is a member of itself. Is 1 a set? Are all numbers sets?

 

[tiny-tim]

Hi Unit!

Your first definition is not a definition …

you can't have the thing you're defining on both sides of the equation.

(and 1 is not in 1 … 1 is the set {φ,{φ}}, so the only elements in 1 are φ and {φ})

 

[Unit]

Hi tiny-tim! Thanks for your reply!

If it is not a definition, then what is it? I'm not trying to be sarcastic here; I actually don't know. Would it qualify as a recursive definition?

Also, is φ the empty set? I have only seen it written as Ø.

I have never heard of defining 1 as {φ, {φ}}! I'm assuming 0 = φ, so 1 is {φ} U {{φ}} = {φ, {φ}}. This is the empty set and the set containing the empty set. Now, there are 2 elements in total ... so would 2 = {φ, {φ}, {φ, {φ}}}?

 

[tiny-tim]

Hi Unit!

If it is not a definition, then what is it? I'm not trying to be sarcastic here; I actually don't know. Would it qualify as a recursive definition?

I suppose you could call it an implicit definition, but the Russell paradox really presupposes that everything is constructed explicitly.

Also, is φ the empty set? I have only seen it written as Ø.

I have never heard of defining 1 as {φ, {φ}}! I'm assuming 0 = φ, so 1 is {φ} U {{φ}} = {φ, {φ}}. This is the empty set and the set containing the empty set. Now, there are 2 elements in total ... so would 2 = {φ, {φ}, {φ, {φ}}}?

ah, I only had a φ to hand (and not a Ø), so I used that.

yes, 2 would be {φ, {φ}, {φ, {φ}}} … that's the standard Peano construction for numbers (except I might be one out … maybe that's 3, and maybe 1 is just {φ}? )

 

[Office_Shredder]

You are off by one. n should have n elements in it. So 0 is the empty set, 1 is the set containing the empty set, etc.