Is the following set equal to the empty set?

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  • #1
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Main Question or Discussion Point

Is the following set equal to the empty set??

A={x:[tex] x\in A\Longrightarrow y\in A ,x\neq y[/tex]},if yes prove it ,if not disproved it
 

Answers and Replies

  • #2
283
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This doesn't make sense. For one thing any thing that is not in A will be in A.
 
  • #3
102
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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??
 
  • #4
147
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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.
 
  • #5
CRGreathouse
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A={x:[tex] x\in A\Longrightarrow y\in A ,x\neq y[/tex]}
This is just
[tex]\forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)[/tex]
which is
[tex]\forall x\;\; x\in A\Longrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)[/tex]
[tex]\left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)\Longrightarrow\forall x\;\; x\in A[/tex]
which is
[tex]\forall x\;\; x\in A\Longrightarrow \left(\forall y\neq x\;\; y\in A\right)[/tex]
[tex]\forall x\;\; \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)\Longrightarrow x\in A[/tex]
which is
[tex]\forall x\;\; x\in A\Longrightarrow \left(\forall y\neq x\;\; y\in A\right)[/tex]
[tex]\forall x\;\; x\in A[/tex]
which is
[tex]\forall x\forall y\neq x\;\; y\in A[/tex]
[tex]\forall x\;\; x\in A[/tex]
which is
[tex]A=\mathcal{U}[/tex]

So if you have a universal set, A is it; if not, the definition is ill-defined.
Edit: What Preno said.
 
  • #6
102
1


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

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: [tex]\forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)[/tex].

DO you actually mean:[tex]x\in A\Longleftrightarrow(x\in A\Longrightarrow\forall y(y\neq x\wedge y\in A))[/tex].

If yes, how did you get that?
 
  • #7
CRGreathouse
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I'm just expanding the definition of set-builder notation.
 
  • #8
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expanding set builder notation is:

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

Now how from the above you get :


[tex]\forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right)[/tex].
 
  • #9
CRGreathouse
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expanding set builder notation is:

[tex] x\in A\Longleftrightarrow[(x\in A\Longrightarrow y\in A)\wedge x\neq y][/tex]
Wait, you really meant
[tex](x\in A\Longrightarrow y\in A)\wedge x\neq y[/tex]
in your original formulation when you wrote
[tex]x\in A\Longrightarrow y\in A ,x\neq y[/tex]?

That's very different!
 
  • #10
102
1


yes

A= { [tex]x: (x\in A\Longrightarrow y\in A)\wedge x\neq y[/tex]}

sorry
 
  • #11
CRGreathouse
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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.
 
  • #12
102
1


the whole sentence
 

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