1.

A={x:x∉x}

Is A a subset of itself?

But my question is:

Let there be a set M such that:

M={1,2,3,4}

The now, one asks if M is a subset of itself. Most probably he would hear a know but,

since M={1,2,3,4} and again I write here M={1,2,3,4}, {1,2,3,4} can be replaced with M.

Thus,

M={1,2,3,4}
M={M}

Thus all sets are subsets of themselves. Now, there is no set x such that x∉x and thus there is no set A where A={x:x∉x}. So where is the paradox?

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disregardthat
You probably mean "I write here M = {{1,2,3,4},{1,2,3,4}}", but as you can see this does not reduce to M.

Furthermore, you have confused "being a subset" with "being an element of" here. Every set is trivially a subset of itself. Also, the axiom of regularity will not allow any set to be an element of itself. But russell's paradox is not conflicting with the axiom of regularity, but rather the axiom of replacement. If you allow unrestricted replacement and insist on the axiom of regularity, russell's paradox will still provide a contradiction!

The paradox is that X = {x | x∉x} is a set if you allow unrestricted replacement, but neither "X is an element of X" or "X is not an element of X" can be true.

In addition, if we say (falsely), as you have "proved", that every set is an element of itself, russell's paradox will even still provide a contradiction by the same argument. X will be the set of all sets in this case (of course still allowing unrestricted replacement), but "X is an element of itself" will not be true, as this implies that X is not an element of itself.

You probably mean "I write here M = {{1,2,3,4},{1,2,3,4}}", but as you can see this does not reduce to M.

Furthermore, you have confused "being a subset" with "being an element of" here. Every set is trivially a subset of itself. Also, the axiom of regularity will not allow any set to be an element of itself. But russell's paradox is not conflicting with the axiom of regularity, but rather the axiom of replacement. If you allow unrestricted replacement and insist on the axiom of regularity, russell's paradox will still provide a contradiction!

The paradox is that X = {x | x∉x} is a set if you allow unrestricted replacement, but neither "X is an element of X" or "X is not an element of X" can be true.

In addition, if we say (falsely), as you have "proved", that every set is an element of itself, russell's paradox will even still provide a contradiction by the same argument. X will be the set of all sets in this case (of course still allowing unrestricted replacement), but "X is an element of itself" will not be true, as this implies that X is not an element of itself.

Ok. So there is no solution? So there is no such set A where A={x:x∉x}?

disregardthat
Unrestricted replacement implies there is such a set, and we don't use unrestricted replacement in ZFC. So there is no such set in ZFC. This is a paradox in the "naive set theory".

I think your major confusion is in saying that a set cannot be a member of itself.

The now, one asks if M is a subset of itself. Most probably he would hear a know but,

since M={1,2,3,4} and again I write here M={1,2,3,4}, {1,2,3,4} can be replaced with M.

Thus,

M={1,2,3,4}
M={M}

Thus all sets are subsets of themselves. Now, there is no set x such that x∉x and thus there is no set A where A={x:x∉x}. So where is the paradox?

First, x and A are different things, so it's best if you call x a set and A a class. All sets are classes (you can define a set as being the collection of all sets that are members of x), and our intuition may prompt us to think that it also works the other way around. Russell's paradox comes into play when we assume the latter.

Let us refer to x as being "set variables", which can be quantified using predicate calculus, etc. Class variables, on the other hand, say something about all possible sets that meet a particular condition. For example, for all sets x, x is an element of the Russell class iff x is not an element of itself.

Now - we need two rules in order to prove Russell's paradox:
the fact that $\neg$($\varphi$$\leftrightarrow$$\neg$$\varphi$)
which is a theorem of classical propositional logic
and a standard axiom of logic,
$\forall$x $\varphi$x $\rightarrow$ $\varphi$y
which should be read "if it's always the case that $\varphi$, then we can substitute y for x in $\varphi$ ($\varphi$ is a well-formed formula that says something about x), and the resulting well-formed formula will also be true with the substitution.

Now, by definition of the Russell class, and using the axiom of Extensionality,
$\forall$x, ( x $\in$ Russell $\Leftrightarrow$ x $\notin$ x ).
Using the above-described quantification theorem of predicate logic, we infer that
Russell $\in$ Russell $\Leftrightarrow$ Russell $\notin$ Russell

But also notice that this contradiction only holds if we either assume that Russell is a set (in which it would be a member of the universal set automatically) or that Russell is a member of the universe. Either way, Russell class's membership of the universe implies this contradiction, so it must be the case that the Russell class doesn't exist.

A={x:x∉x}

Is A a subset of itself?

No, Russell's paradox asks if A is a member of itself ($A \in A$), not a subset of itself ($A \subseteq A$). Your entire question rests solely in the confusion between element and subset; in fact, all sets are, trivially, subsets of themselves.

No, Russell's paradox asks if A is a member of itself ($A \in A$), not a subset of itself ($A \subseteq A$). Your entire question rests solely in the confusion between element and subset; in fact, all sets are, trivially, subsets of themselves.

Well, actually { x | x $\subseteq$ x }
is equal to the Universal set, because x e. { x | x = x } iff x = x. x = x iff x $\subseteq$ x (simply because both are true). { x | x $\subseteq$ x } = { x | x = x }

And in ZFC, A = B -> A e/ B, so this set you talk about there wouldn't be an element of the universe either :). And the described classes (universe, russell set, and x $\subseteq$ x) are all equal to one another (although you can prove Russell's paradox, as described above, by using only Extensionality).