Proving or Disproving a Statement in Set Notation

[amilapsn]

Homework Statement

Prove or disprove the following
(i) $\forall a\in\mathbb{R}[(\forall \epsilon>0,a<\epsilon)\Leftrightarrow a\leq 0]$

2. The attempt at a solution
Can't we disprove the above statement by telling $a\leq 0 \nRightarrow (\forall \epsilon>0,a<\epsilon)$ through a counter example like $a\leq 0 \Rightarrow (\epsilon=0,a\leq \epsilon)$ or something?

 

[PeroK]

Can you write out in words what the proposition is saying?

What would you have to show to prove it? Hint: there would be two parts to proving it.

Can you write down what property a countereaxmple would have? Hint: it could have one of two properties.

Does it hold for a = 0? Does it hold for a = -1? Does it hold for a = 1?

 

[amilapsn]

The proposition in words:
for all a belongs to real number set , for all $\epsilon$ >0 and a<$\epsilon$ if and only if $a\leq 0$ .

I have to show both $\forall \epsilon >0,a<\epsilon\Rightarrow a\leq 0$ and $a\leq 0\Rightarrow \forall \epsilon >0,a<0$ to prove the proposition.

A counter example should disprove the proposition.

 

[PeroK]

The proposition in words:
for all a belongs to real number set , for all $\epsilon$ >0 and a<$\epsilon$ if and only if $a\leq 0$ .

I have to show both $\forall \epsilon >0,a<\epsilon\Rightarrow a\leq 0$ and $a\leq 0\Rightarrow \forall \epsilon >0,a<0$ to prove the proposition.

A counter example should disprove the proposition.

That's good.

What about a = 0, 1, -1? Does the proposition hold for these values of a?

 

[amilapsn]

The proposition holds for a=-1,0. But it doesn't hold for a=1.

 

[PeroK]

The proposition holds for a=-1,0. But it doesn't hold for a=1.

Why does it fail for a = 1?

 

[amilapsn]

Why does it fail for a = 1?

Because a=1 is not less than for all $\epsilon>0$

 

[PeroK]

Because a=1 is not less than for all $\epsilon>0$

Take a step back. We have a proposition:

$\forall a \ \ A \Leftrightarrow B$

That's means that (if the proposition holds) then for each a we have either: A(a) true and B(a) true; or A(a) false and B(a) false.

For a = 1, what can you say about A(1) and B(1)?

 

[amilapsn]

I see. The proposition holds for a=1 too, because A(1) false and B(1) false. Thanks...
Thank you for showing me the better way to look at the question.

 

[amilapsn]

Then the proposition is true for all a, so that we can't disprove it. We have to prove it. Thanks again @PeroK

 

[PeroK]

I see. The proposition holds for a=1 too, because A(1) false and B(1) false. Thanks...
Thank you for showing me the better way to look at the question.

Also, when I first asked you to describe the proposition in words, you could have said:

The proposition states that:

"Any real number is less than or equal to 0 iff it is less than every positive number".

Put like that, it's clear that the proposition holds.

 

[amilapsn]

Yeah, it's really clearer...

 

[PeroK]

Then the proposition is true for all a, so that we can't disprove it. We have to prove it. Thanks again @PeroK

Here's a tip. This is something I do when dealing with propositions and logic:

I use "true" and "false" to relate to individual statements. E.g. $a > 0$ can be true or false.

And, I say a proposition "holds" or "fails". E.g. the proposition holds for a = 1.

 

[amilapsn]

Just now I felt what is called as "Enlightenment..."