Post History
#8: Post edited
by
MathPhysics
·
2020-10-09T09:43:06Z (over 3 years ago)
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
Thus, a set of all sets cannot exist in such an axiomatic system.$$\begin{CD} A @>a>> B \\\\ @V b V V= @VV c V \\\\ C @>>d> D \end{CD}$$
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
- Thus, a set of all sets cannot exist in such an axiomatic system.
#7: Post edited
by
MathPhysics
·
2020-10-09T09:42:38Z (over 3 years ago)
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
Thus, a set of all sets cannot exist in such an axiomatic system.
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
- Thus, a set of all sets cannot exist in such an axiomatic system.
- $$\begin{CD} A @>a>> B \\\\ @V b V V= @VV c V \\\\ C @>>d> D \end{CD}$$
#6: Post edited
by
MathPhysics
·
2020-10-09T09:29:49Z (over 3 years ago)
$$\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \end{pmatrix}$$
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
- Thus, a set of all sets cannot exist in such an axiomatic system.
#5: Post edited
by
MathPhysics
·
2020-10-09T09:29:06Z (over 3 years ago)
$$\require{AMScd}\begin{CD} A @>a>> B \\\\ @V b V V= @VV c V \\\\ C @>>d> D \end{CD}$$
- $$\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \end{pmatrix}$$
#4: Post edited
by
MathPhysics
·
2020-10-09T09:27:57Z (over 3 years ago)
$$\overbrace{x+y+z}$$Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:$$U= \\{ x\in V \mid x \not \in x \\}.$$Now, there are two cases:- $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;- $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.Thus, a set of all sets cannot exist in such an axiomatic system.
- $$\require{AMScd}\begin{CD} A @>a>> B \\\\ @V b V V= @VV c V \\\\ C @>>d> D \end{CD}$$
#3: Post edited
by
MathPhysics
·
2020-10-09T09:26:01Z (over 3 years ago)
$$\require{AMScd}\begin{CD} A @>a>> B \\\\ @V b V V= @VV c V \\\\ C @>>d> D \end{CD}$$- $$\overbrace{x+y+z}$$
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
- Thus, a set of all sets cannot exist in such an axiomatic system.
- $$\overbrace{x+y+z}$$
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
- Thus, a set of all sets cannot exist in such an axiomatic system.
#2: Post edited
by
MathPhysics
·
2020-10-09T09:25:06Z (over 3 years ago)
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
- Thus, a set of all sets cannot exist in such an axiomatic system.
- $$\require{AMScd}\begin{CD} A @>a>> B \\\\ @V b V V= @VV c V \\\\ C @>>d> D \end{CD}$$
- $$\overbrace{x+y+z}$$
- Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
- $$U= \\{ x\in V \mid x \not \in x \\}.$$
- Now, there are two cases:
- - $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- - $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
- Thus, a set of all sets cannot exist in such an axiomatic system.
#1: Initial revision
by
MathPhysics
·
2020-10-09T08:45:20Z (over 3 years ago)
Let $V$ be a set of all sets. According to the **Axiom Schema of Comprehension**, we can have the following set:
$$U= \\{ x\in V \mid x \not \in x \\}.$$
Now, there are two cases:
- $U \in U$, which, according to the definition of the set $U$, implies that $U \not \in U$, which is a contradiction;
- $U \not \in U$, which, according to the definition of the set $U$, implies that $U \in U$, which is a contradiction.
Thus, a set of all sets cannot exist in such an axiomatic system.