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The simplest of the logical paradoxes is Russell's paradox, which can be described as follows:

Let $S$ denote the set of all sets that are not elements of themselves. Is $S$ an element of itself?

  • Well, if $S$ is an element of $S$, then - by the very definition of $S$ - $S$ is not an element of $S$.

  • If $S$ is not an element of $S$, then (again, because of the way $S$ is defined) $S$ is an element of $S$.

Thus, we have proven that $S$ is an element of $S$ if and only if $S$ is not an element of $S$ - a contradiction of the most fundamental sort.

Source: Set Theory by Charles C. Pinter, p.5 [Formatting not withstanding].

From the above explanation, reductio ad absurdum can be applied to the statement '$S$ is a set of all sets that are not elements of themselves.'as follows: $$\begin{align} S \in S &\Rightarrow S \notin S\\ &\Rightarrow S\in S\\ &\therefore S\in S \Leftrightarrow S\notin S \end{align}$$

Since the statement leads to a contradictin, $\color{purple}{ S\in S \Leftrightarrow S\notin S}$, doesn't the statement '$S$ is a set of all sets that are not elements of themselves.' hold true?

[Edit Ahh. I'm starting to understand it a bit more]

The quote "Let $S$ denote the set of all sets that are not elements of themselves. Is $S$ an element of itself?" can be rephrased as 'If S denote the set of all sets that are not elements of themselves, then $S$ an element of itself'.

'S is a set of all sets that are not elements of themselves" is both the hypothesis and the statement. Kevin Houston, page 64:

"Statement A is called the hyothesis or assumption" - Susanna S. Epp, page 390: "Euclidean Algorithm 1. Let A and B be integers with $A≥B≥0$."...recall that we assumed $A≥B≥0$

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  • $\begingroup$ Other than creating broken CSS, eyestrain and the feeling that I'm looking at a drawing of a child, what point is there to the colors? $\endgroup$ – Asaf Karagila May 30 '16 at 8:37
  • $\begingroup$ @AsafKaragila The colors are there for increasing the readability of my question. :) By the way, what does CSS mean? $\endgroup$ – buzzee May 30 '16 at 8:41
  • $\begingroup$ Cascaded Style Sheets. The design of the website. Currently broken, courtesy of the colored text overflowing from the question box. $\endgroup$ – Asaf Karagila May 30 '16 at 8:42
  • $\begingroup$ @buzzee What is the point of pasting page 43 of your textbook? $\endgroup$ – 5xum May 30 '16 at 8:46
  • $\begingroup$ @5xum I compared it with answers. $\endgroup$ – buzzee May 30 '16 at 8:53
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NO: the statement

"there exists the set $S$ of all sets that are not elements of themselves"

has been proved to be false.

The "logical analysis" of Russell's Paradox starts from the seemingly harmless principle [called: unrestricted Comprehension Axiom] that:

for any formula $φ(x)$ containing $x$ as a free variable, there will exist the set $\{ x \mid φ(x) \}$.

Russell manufactured his paradox considering as $\varphi$ the predicate : $\lnot (x \in x)$.

Thus, according to the above principle, we are licensed to make the assumption:

$\exists S \ \forall x \ (x \in S \leftrightarrow \lnot (x \in x))$ --- (*)

Applying inference rules, and instantiating the universal quantifier with $S$ we get:

$S \in S \leftrightarrow \lnot (S \in S)$.

This is the contradiction.

Thus, applying reductio to the initial assumption (*), we conclude with:

$\lnot \exists S \ \forall x \ (x \in S \leftrightarrow \lnot (x \in x))$.

And so we have to conclude that the unrestricted Comprehension Axiom is not a "sound" principle: we have to leave it, or to "restrict" in a suitable way its application.

See Axiom schema of specification for a way to "fix" it.

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You're misreading the structure of the argument. The structure is:

  • Suppose $S$ has the property $\forall x(x \in S \iff x \notin x)$.

  • Then $S \in S \iff S \notin S$, a contradiction.

  • Therefore, the assumption that $S$ has the property $\forall x(x \in S \iff x \notin x)$ was absurd.

  • In other words, no set $S$ has the property $\forall x(x \in S \iff x \notin x).$

  • Symbolically:

$$\forall S( \neg \forall x(x \in S \iff x \notin x))$$

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    $\begingroup$ I have tried putting it as "It is illogical to claim that there exists a widget that doodles all those and only those widgets that do not doodle themselves." $\endgroup$ – DanielWainfleet May 30 '16 at 18:53

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