Set theory

My log for studying set theory

This pages contains log for my self-study on set theory.

Chapter 1: Sets

3. The axioms

3.1

Let \(P(x, A, B)\) be the property: \(x \in A, \, x \notin B\). Then this property implies that \(x \in A\). Therefore, \(\{ x \mid x \in A, x \in B \} = \{ x \in A \mid x \in A, x \in B \} = \{ x \in A \mid x \in B \}\). This set exists by the Axiom schema of comprehension.

3.2

Let \(A\) be a set known to exist. Then using property \(P(x) = x \neq x\), we can use Axiom schema of comprehension to say that there exists set \(B = \{ x \in A \mid x \neq x \}\). Then we can easily see that such set is empty.

3.3 (a)

Suppose that there is a set of all sets \(V\). Then using Axiom schema of comprehension, there exists a set \(T = \{ x \in V \mid x \notin x \}\). Now we will show contradiction by saying \(T \notin V\) since this would mean \(T\) is not a set. Suppose \(T \in V\). Suppose \(T \notin T\). Then \(T \in T\), contradiction. Suppose \(T \in T\). Then \(T \notin T\) must hold, again contradiction. Thus \(T \notin V\).

3.3 (b)

This is easily proven because if there is some set \(A\) where \(x \in A\) for all \(x\), this would mean \(A\) is set of all sets.

3.4

This can be proven by doing axiom of schema comprehension and then doing axiom of union.

3.5

This can be proven by doing Axiom of pair and union.

3.6

We will use the hint. we can clearly see that \(Y \in p(x)\) since \(Y\) by definition is a subset of \(X\). Then we can derive contradiction by checking if \(Y \in Y\). This logic is similar to the previous proofs that showed that “set of all sets” does not exist.

3.7

Consult link.

4. Elementary operations on sets

4.4

Suppose that \(A^c\) exists. Then by axiom of union, the union of \(A\) and \(A^c\) exists. But this is set of all sets. Contradiction.

4.6