1. Originally Posted by Sampras
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So you want to show: $\displaystyle \exists \epsilon >0, \ \text{s.t.} \ \forall \delta >0, \ |x| < \delta \implies |f(x)-f(0)| \geq \epsilon$
But you wrote the above definition for discontinuity which is wrong

2. Originally Posted by xalk
But you wrote the above definition for discontinuity which is wrong
$\displaystyle \exists \epsilon >0, \ \text{s.t.} \ \forall \delta >0, \exists x \ \text{s.t.} \ |x| < \delta \implies |f(x)-f(0)| \geq \epsilon$

3. Okay, reading back on the previous posts it appears my intial proof wasn't correct.

Here's my new version:

Let $\displaystyle x_n=\frac{2}{\pi(2n+1)} \ \ \ \forall n \in \mathbb{Z}^+$

$\displaystyle \Rightarrow f(x_n)=3 \sin \left( \frac{(2n+1)\pi}{2} \right)= \begin{cases} 3 \ \ \ \ n \in 2 \mathbb{Z}^+\\ -3 \ \ \ \ n \in \mathbb{Z}^+-2\mathbb{Z}^+ \end{cases}$

$\displaystyle \delta>0 \Rightarrow \exists x_k \ s.t \ |0-x_k|< \delta \Rightarrow |f(0)-f(x_k)|=|0 \pm 3|=3$

hence pick $\displaystyle \epsilon:= \frac{3}{2}$

When $\displaystyle \delta>0$ and $\displaystyle x_n=\frac{2}{\pi(2n+1)}$:

$\displaystyle |0-x_k|<\delta \Rightarrow |f(0)-f(x_k)|=|0 \pm 3|=3>\epsilon$ as required.

Is that better?

4. Originally Posted by Sampras
$\displaystyle \exists \epsilon >0, \ \text{s.t.} \ \forall \delta >0, \exists x \ \text{s.t.} \ |x| < \delta \implies |f(x)-f(0)| \geq \epsilon$
.

But this definition is agian wrong.

The correct one is:

$\displaystyle \exists\epsilon>0$ and $\displaystyle \forall\delta>0\Longrightarrow\exists x$ such that ,|x|<δ AND |f(x)-f(o)|$\displaystyle \geq\epsilon$.

Which in words means:
After we have chosen ε>0 ,and given δ>0 we must find an x such that:

|x|<δ AND |f(x)-f(0)|$\displaystyle \geq\epsilon$

5. Originally Posted by xalk
But this definition is agian wrong.
Once again xalk, you had best be sure you understand what you are talking about.
Here is the symbolic definition of the function $\displaystyle f$ being continuous at $\displaystyle x=a$:
$\displaystyle \left( {\forall \varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left( {\forall x} \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| < \varepsilon } \right]$.

Do you understand how to negate quantifiers?

The negation of that statement is:
$\displaystyle \left( {\exists \varepsilon > 0} \right)\left( {\forall \delta > 0} \right)\left( {\exists b} \right)\left[ {\left| {b - a} \right| < \delta \wedge \left| {f(b) - f(a)} \right| \geqslant \varepsilon } \right]$
Please notice that there is no implication in that negation.

6. Originally Posted by Plato
Once again xalk, you had best be sure you understand what you are talking about.
Here is the symbolic definition of the function $\displaystyle f$ being continuous at $\displaystyle x=a$:
$\displaystyle \left( {\forall \varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left( {\forall x} \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| < \varepsilon } \right]$.

Do you understand how to negate quantifiers?

The negation of that statement is:
$\displaystyle \left( {\exists \varepsilon > 0} \right)\left( {\forall \delta > 0} \right)\left( {\exists b} \right)\left[ {\left| {b - a} \right| < \delta \wedge \left| {f(b) - f(a)} \right| \geqslant \varepsilon } \right]$
Please notice that there is no implication in that negation.

The definition of continuity at a point a is :

$\displaystyle \forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\we dge\forall x( |x-a|<\delta\Longrightarrow |f(x)-f(a)|<\epsilon))]$

And the negation of that definition is:

$\displaystyle \exists\epsilon[\epsilon>0\wedge\forall\delta(\delta>0\Longrightar row\exists x( |x-a|<\delta\wedge |f(x)-f(a)|\geq\epsilon))]$.

You will find the definition and its negation in ANGELO'S MARGARIS book, titled FIRST ORDER MATHEMATICAL LOGIC ,on pages 45 and 46 ,exercise 1(f).

Now if you want me to prove how the negation of the definition of continuity is implied ,i can do that.

This a good example showing what the value is of a correct statement in quantifier form.

Both statements have implications

7. Originally Posted by xalk
The definition of continuity at a point a is :

$\displaystyle \forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\we dge\forall x( |x-a|<\delta\Longrightarrow |f(x)-f(a)|<\epsilon))]$

And the negation of that definition is:

$\displaystyle \exists\epsilon[\epsilon>0\wedge\forall\delta(\delta>0\Longrightar row\exists x( |x-a|<\delta\wedge |f(x)-f(a)|\geq\epsilon))]$.

You will find the definition and its negation in ANGELO'S MARGARIS book, titled FIRST ORDER MATHEMATICAL LOGIC ,on pages 45 and 46 ,exercise 1(f).

Now if you want me to prove how the negation of the definition of continuity is implied ,i can do that.

This a good example showing what the value is of a correct statement in quantifier form.

Both statements have implications
That's exactly what Plato said, isn't it? The "implication" here is at an entirely different place.

8. Originally Posted by xalk
The definition of continuity at a point a is :

$\displaystyle \forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\we dge\forall x( |x-a|<\delta\Longrightarrow |f(x)-f(a)|<\epsilon))]$

And the negation of that definition is:

$\displaystyle \exists\epsilon[\epsilon>0\wedge\forall\delta(\delta>0\Longrightar row\exists x( |x-a|<\delta\wedge |f(x)-f(a)|\geq\epsilon))]$.

You will find the definition and its negation in ANGELO'S MARGARIS book, titled FIRST ORDER MATHEMATICAL LOGIC ,on pages 45 and 46 ,exercise 1(f).

Now if you want me to prove how the negation of the definition of continuity is implied ,i can do that.

This a good example showing what the value is of a correct statement in quantifier form.

Both statements have implications
My perspective on your argument is as follows:

Let A be a well-formed formula in the first order language.

(1) $\displaystyle \forall\delta(\delta>0\rightarrow A)$.
(2) $\displaystyle (\forall\delta > 0) A$.
(3) $\displaystyle \forall\delta >0 \rightarrow A$.

(1) and (2) are equivalent, but (1) and (2) are not equivalent to (3). Although $\displaystyle \forall\delta(\delta>0)$ is a well-formed formula (Two place predicate symbol ">" should be defined), (3) is not wff in the first order language.

I think you assumed (1),(2) and (3) are equivalent, since your previous post said,

$\displaystyle \exists\epsilon>0$ and $\displaystyle \forall\delta>0\Longrightarrow\exists x$ such that ,|x|<δ AND |f(x)-f(o)|$\displaystyle \geq\epsilon$.

But it is not true. This is why implication is not necessary if we use form (2).

9. Originally Posted by aliceinwonderland
My perspective on your argument is as follows:

Let A be a well-formed formula in the first order language.

(1) $\displaystyle \forall\delta(\delta>0\rightarrow A)$.
(2) $\displaystyle (\forall\delta > 0) A$.
(3) $\displaystyle \forall\delta >0 \rightarrow A$.

(1) and (2) are equivalent, but (1) and (2) are not equivalent to (3). Although $\displaystyle \forall\delta(\delta>0)$ is a well-formed formula (Two place predicate symbol ">" should be defined), (3) is not wff in the first order language.

I think you assumed (1),(2) and (3) are equivalent, since your previous post said,

$\displaystyle \exists\epsilon>0$ and $\displaystyle \forall\delta>0\Longrightarrow\exists x$ such that ,|x|<δ AND |f(x)-f(o)|$\displaystyle \geq\epsilon$.

But it is not true. This is why implication is not necessary if we use form (2).

To cut a long story short.

Do you or do you not agree with the formulas written by ANGELO MARGARIS in his book ,FIRST ORDER MATHEMATICAL LOGIC ,pages 45 to 46??

If YOU DO ,there is nothing to be said ,because i do agree with him,except that PLATO'S formulas are wrong.

IF YOU DO NOT, then any arguments should be against ANGELO MARGARIS.

AND I DO APOLOGIZE,if in trying to write those formulas in forms other than their purely symbolical ,i did any mistakes.

This does not mean ,that i cannot respond to your post in more details.

CERTAINLY I CAN ,but somebody will come along and lock the thread ,leaving everybody with simple impressions and loosing the essence of the problem

10. Originally Posted by xalk
Do you or do you not agree with the formulas written by ANGELO MARGARIS in his book ,FIRST ORDER MATHEMATICAL LOGIC ,pages 45 to 46??
If YOU DO ,there is nothing to be said ,because i do agree with him, except that PLATO'S formulas are wrong.
IF YOU DO NOT, then any arguments should be against ANGELO MARGARIS.
Having taught courses from Margaris’ book, even though it was years ago, I would say that I well understand the points he is making.
I think that it is you who do not understand the logic here.

On page 45, the instructions for #1 say “Find a useful denial…”
The statement in part (e) is the symbolic statement for continuity; whereas the statement in part (f) is not.
The negation given on page 46 is exactly the one I gave above without any implication.
This is a denial of a universal giving an existential which does not contain an implication.

11. Originally Posted by Plato
Having taught courses from Margaris’ book, even though it was years ago, I would say that I well understand the points he is making.
I think that it is you who do not understand the logic here.

On page 45, the instructions for #1 say “Find a useful denial…”
The statement in part (e) is the symbolic statement for continuity; whereas the statement in part (f) is not.
The negation given on page 46 is exactly the one I gave above without any implication.
This is a denial of a universal giving an existential which does not contain an implication.

The formula in part (e) of the question in Margari's book is:

$\displaystyle \forall\epsilon\exists\delta\forall x(|x-c|<\delta\Longrightarrow |f(x)-f(c)|<\epsilon)$

The formula in part (f) of the question in the same book is:

$\displaystyle \forall\epsilon[\epsilon>0\Longrightarrow\exists\delta(\delta>0\we dge\forall x( |x-c|<\delta\Longrightarrow |f(x)-f(c)|<\epsilon))]$

Which is the formula that i wrote in my post # 21

Now if the formula (e) is the right one ,what is the formula (f) for.

Besides ,when we start a proof for continuity at a point ,we start with the assumption:

Let ε>0, a feature which is part of formula (f) and not of formula (e).

IN A FORMAL PROOF for continuity at a point ,only formula (f) is applicable and not formula (e).

HENCE the right formula is (f) and its negation is the formula that i wrote in my post #21

However if you still insist that the right formula is (e) ,you have to prove that .

In a formal proof ,where everything is written down one can easily see that the right formula is (f),whilst in an ordinary mathematical proof the whole issue can be confused.

For example in formaly proving that the simple function,f(x) =2x+3 is continuous at any point c we can only use formula (f)

12. This thread is getting out of hand, so I'm going to close it. Remember everyone, SHOUTING and typing in bold doesn't make what you say correct, it's just impolite. Let's all be courteous to each other.

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