Is the following set equal to the empty set?

[poutsos.A]

Is the following set equal to the empty set??

A={x:$$x\in A\Longrightarrow y\in A ,x\neq y$$},if yes prove it ,if not disproved it

 

[Focus]

This doesn't make sense. For one thing any thing that is not in A will be in A.

 

[poutsos.A]

A Is a set of x elements in a such way that if x belongs to A THEN ANY y different from x belongs to A.

Doesn't that make sense??

 

[Preno]

Well, if A existed, it would be the universal set (because of what Focus said). But there is no universal set. So A is not a set.

 

[CRGreathouse]

A={x:$$x\in A\Longrightarrow y\in A ,x\neq y$$}

This is just
$$\forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)$$
which is
$$\forall x\;\; x\in A\Longrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)$$
$$\left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)\Longrightarrow\forall x\;\; x\in A$$
which is
$$\forall x\;\; x\in A\Longrightarrow \left(\forall y\neq x\;\; y\in A\right)$$
$$\forall x\;\; \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)\Longrightarrow x\in A$$
which is
$$\forall x\;\; x\in A\Longrightarrow \left(\forall y\neq x\;\; y\in A\right)$$
$$\forall x\;\; x\in A$$
which is
$$\forall x\forall y\neq x\;\; y\in A$$
$$\forall x\;\; x\in A$$
which is
$$A=\mathcal{U}$$

So if you have a universal set, A is it; if not, the definition is ill-defined.
Edit: What Preno said.

 

[poutsos.A]

This is just
$$\forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)$$
which is
$$\forall x\;\; x\in A\Longrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)$$
$$\left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)\Longrightarrow\forall x\;\; x\in A$$
which is
$$\forall x\;\; x\in A\Longrightarrow \left(\forall y\neq x\;\; y\in A\right)$$
$$\forall x\;\; \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)\Longrightarrow x\in A$$
which is
$$\forall x\;\; x\in A\Longrightarrow \left(\forall y\neq x\;\; y\in A\right)$$
$$\forall x\;\; x\in A$$
which is
$$\forall x\forall y\neq x\;\; y\in A$$
$$\forall x\;\; x\in A$$
which is
$$A=\mathcal{U}$$

So if you have a universal set, A is it; if not, the definition is ill-defined.
Edit: What Preno said.

.

LETS take it line by line:

1st line you have written: $$\forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)$$.

DO you actually mean:$$x\in A\Longleftrightarrow(x\in A\Longrightarrow\forall y(y\neq x\wedge y\in A))$$.

If yes, how did you get that?

 

[CRGreathouse]

I'm just expanding the definition of set-builder notation.

 

[poutsos.A]

expanding set builder notation is:

$$x\in A\Longleftrightarrow[(x\in A\Longrightarrow y\in A)\wedge x\neq y]$$

Now how from the above you get :


$$\forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)$$.

 

[CRGreathouse]

expanding set builder notation is:

$$x\in A\Longleftrightarrow[(x\in A\Longrightarrow y\in A)\wedge x\neq y]$$

Wait, you really meant
$$(x\in A\Longrightarrow y\in A)\wedge x\neq y$$
in your original formulation when you wrote
$$x\in A\Longrightarrow y\in A ,x\neq y$$?

That's very different!

 

[poutsos.A]

yes

A= { $$x: (x\in A\Longrightarrow y\in A)\wedge x\neq y$$}

sorry

 

[CRGreathouse]

So your definition is given in terms of the unbound variable y? I was assuming in my translation that you intended for the definition to be a sentence.

 

[poutsos.A]

the whole sentence