0
votes
2answers
82 views

Russell's paradox with bounded comprehension

Consider the set $S = \{A, \varnothing\}$ and define $A = \{x \in S|x \not\in x\}$; this is the same as Russell's paradox except with bounded comprehension, ie $A\in A\iff A\not\in A$. I think the ...
1
vote
1answer
80 views

Why $\bigcap \emptyset $ isn't defined? [duplicate]

Let us define: $\bigcap \emptyset = \{ x|\forall A(A \in \emptyset \Rightarrow x \in A)\} $ I understood that this cannot be defined. Somehow it enables Russell's paradox to exist. Why is that?
2
votes
2answers
69 views

Variant on Russell's paradox: show $B = \varnothing$

Let $X$ be a set and $R$ a relationship on $X$. Define $N = \{x \in X\mid(x, x) \notin R\}$. Let $$B =\{b \in X\mid(\forall n \in N)(b\,R\,n) \land (\forall n \notin N)(\neg b\,R\,n)\}\;.$$ ...
12
votes
5answers
614 views

Why cannot a set be its own element?

When I study Topology, I met with a problem. On my book, it says 'we cannot admit that there exists a set whose members are all the topological spaces. That will lead to a logical contradiction, that ...
4
votes
3answers
879 views

The set of all infinite binary sequences

Suppose that we have the set $S$ of all possible infinite binary sequences $s_i$ (a sequence is simply an ordered set): $$S=\{s_1,s_2,s_3,\ldots \}$$ where the sequences $s_i$ are like ...
-2
votes
2answers
273 views

Does a complement contain itself?

I think it is safe to assume many sets do not contain their complements. {1, 2} for example. Now, by the definition of complement, that would mean the complement would have to contain itself. This ...
3
votes
3answers
303 views

Why is the hypergame not simply well founded?

According to Cameron, the hypergame paradox proceeds as follows: A game is considered as well founded if ANY play of the game ends in a finite number of moves. A hypergame is where the first player ...
2
votes
2answers
189 views

Can one come to prove Cantor's theorem (existence of higher degree of infinities) FROM Russell's paradox?

I have been thinking about this: One can arrive at Russell's paradox from Cantor's argument, but can we go the other way round, i.e., can we prove Cantor's diagonal argument(often referred to as ...
24
votes
3answers
4k views

difference between class, set , family and collection

In school I have always seen sets. But I was watching a video the other day about functors and they started talking about any set being a collection but not vice-versa and I also heard people talking ...
2
votes
2answers
388 views

An unwanted property of the set $T=\{x,\{x\} \}$

Let $T$ be $T=\{x,\{x\},y \}$ and let $f:A\rightarrow T, \ f(a):=x$, where $A=\{a\}$. Define $B=\{x,y \}$. Now a weird thing happens: We should have that $x \in f(f^{-1}(B))$ by construciton of $f$, ...
4
votes
1answer
185 views

How is the Cantor's paradox resolved in the ZFC system?

How is the Cantor's paradox resolved in the ZFC system? Thanks.