# Math Help - Sets question.

1. ## Sets question.

Let a1, d2, a3, ... an, ... be a sequence of real numbers. The sequence converges to a real number a if:

[see attachment]

Work out the negation of the sentence, ie a definition of what it means to say that the sequence does not converge to a.

2. Hi
The negation for a " $\forall\ \exists\ \forall\ PROPOSITION$ " formula will be " $\exists\ \forall\ \exists\ \text{no}(PROPOSITION)$ "

$\exists\epsilon >0\ \forall N\in\mathbb{N}\ \exists n>N\ \text{s.t.}\ |a_n-a|\geq\epsilon$

3. Originally Posted by clic-clac
Hi
The negation for a " $\forall\ \exists\ \forall\ PROPOSITION$ " formula will be " $\exists\ \forall\ \exists\ \text{no}(PROPOSITION)$ "

$\exists\epsilon >0\ \forall N\in\mathbb{N}\ \exists n>N\ \text{s.t.}\ |a_n-a|\geq\epsilon$
clic-clac can you prove what you have just written down?? i.e

~ $\forall\epsilon[\epsilon>0\rightarrow\exists k(k\in N\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))]$ is equivalent to:

......... $\exists\epsilon[\epsilon>0\wedge\forall k(k\in N\rightarrow\exists n(n\geq k\wedge |a_n-a|\geq\epsilon))]$.................................................. ................................

4. Hi
Well same thing. Since $\neg( A\Rightarrow B)\equiv A\wedge \neg B$ and $\neg (A\wedge B)\equiv \neg A\vee\neg B\equiv A\Rightarrow\neg B,$ we have:

$\neg(\forall\epsilon[\epsilon>0\rightarrow\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))])$
$\equiv\exists\epsilon[\epsilon>0\wedge\neg(\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$
$\equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$
$\equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\exists n(n\geq k\wedge |a_n-a|>\epsilon))]$

5. Originally Posted by clic-clac
Hi
Well same thing. Since $\neg( A\Rightarrow B)\equiv A\wedge \neg B$ and $\neg (A\wedge B)\equiv \neg A\vee\neg B\equiv A\Rightarrow\neg B,$ we have:

$\neg(\forall\epsilon[\epsilon>0\rightarrow\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))])$
$\equiv\exists\epsilon[\epsilon>0\wedge\neg(\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$
$\equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$
$\equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\exists n(n\geq k\wedge |a_n-a|>\epsilon))]$
Good proof ,but from the 2nd step on wards i tried to find suitable laws of logic to justify each step and i could not.

O.k for the first step since there is the theorem in predicate calculus to account for that,i.e

..............~ $\forall u P\Longleftrightarrow\exists u\neg P$.......................................

But from there i could not find any laws to justify each step ,since you kept the quantification symbols at each step and you did work inside the formula ,without getting rid of the quantification symbols.

There are laws in predicate calculus for removing and introducing the quantification symbols and that in combination with the laws of propositional calculus can give you a proof

6. The only things I used are what I said:
and
You don't have to get rid of the quantification symbols, just replace them by the other one.

For instance, from second step to third:

$\exists\epsilon[\epsilon>0\wedge\neg(\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$

$\equiv\exists\epsilon[\epsilon>0\wedge\forall k\neg(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$

But
$\neg(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))$ has the form $\neg(A\wedge B)$ so it's the same than $k\in \mathbb{N}\Rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))$

we replace it in the whole formula:

$\equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$

7. Originally Posted by clic-clac
You don't have to get rid of the quantification symbols, just replace them by the other one.

But there is no law of logic that allow you to do that ,at least as far as i know.

If however you happen to know of any such law ,mention it so that we could see how that law is derived from the axioms of the predicate calculus

8. Originally Posted by archidi
But there is no law of logic that allow you to do that ,at least as far as i know.
archidi;267897, I have tried to stay out of this thread but now I must make a comment.
archidi;267897 do you know any real logic beyond the level of SCHAUM’S OUTLINE?
I can suggest some much better texts for you to study.
Symbolic Logic by Irving Copi is still the most teachable textbook.
It will give you a deeper understanding of quantifier theory.

The other classics are two texts by Quine: Methods of logic and Set Theory and its Logic.
Set Theory and Logic
by Stoll is also a classic.

You seem to fancy yourself as a logician, but your postings in this tread belie that notion.

9. Originally Posted by Plato
archidi;267897, I have tried to stay out of this thread but now I must make a comment.
archidi;267897 do you know any real logic beyond the level of SCHAUM’S OUTLINE?
I can suggest some much better texts for you to study.
Symbolic Logic by Irving Copi is still the most teachable textbook.
It will give you a deeper understanding of quantifier theory.

The other classics are two texts by Quine: Methods of logic and Set Theory and its Logic.
Set Theory and Logic by Stoll is also a classic.

You seem to fancy yourself as a logician, but your postings in this tread belie that notion.

Thank you plato for your kind words and advise,but definitely i never said that i fancy my self as a logician.

I simply ask questions when i see something that i think it is wrong and then produce my arguments to substantiate the case .

My dear plato i suppose you remember that when i first joined the forum you where the one i was asking questions.

But i also happen to have a great teacher in logic whose knowledge in logic is internationally recognized.

Under his patient advise and unrepeatable teaching i learned a lot about logic within a very short time.

The books now in my library are:

1) Introduction To Symbolic Logic and its Applications,by Rudolf Camp

2) First Course in Mathematical Logic ,by Patrick Suppes and Shirley Hill

3) Popular Lectures on Mathematical Logic, by Hao Wang

4)Theory of Formal Systems, by RAYMOND M. SMULLYAN

5)Logic as Algebra, by PAUL HALMOS and STEVEN GIVANT

6) Set Theory and Logic ,by ROBERT R. STOLL

7)Logic Computing Machines and Automation,by Alice Mary Hilton

8)Introduction to Elementary Mathematical Logic ,by Abram Aronovich Stolyar

9)Introduction To Logic fifth edition ,by IRVING M. COPI

10)Symbolic Logic fourth edition ,by IRVING M. COPI

11)Foundations without Foundationalism A CASE FOR SECOND ORDER LOGIC ,by Stewart Shapiro

12) Introduction To Logic AND TO THE METHODOLOGY OF DEDUCTIVE SCIENCES ,by Alfred Tarski

13)Introduction To Logic , by PATRICK SUPPES

14) What is Mathematical Logic ,by J N Crossley,C.J.Ash,C.J.Brickhill,J.C.Stillwell,N.H.W illiams

15)FIRST ORDER LOGIC ,by Raymond M.Smullyan

16) FIRST ORDER MATHEMATICAL LOGIC by Angelo Margaris

17)BOOLEAN ALGEBRA and SWITHING CIRCUITS IN SCHAUM'S OUTLINE SERIES.by ELLIOT MENDELSON

18)LOGIC IN SCHAUM'S OUTLINE SERIES by John Nolt and Dennis Rohatyn

plus other books in other languages.

10. Originally Posted by archidi
But i also happen to have a great teacher in logic whose knowledge in logic is internationally recognized.
Well I doubt that first statement. Is your idea of international is limited?
What has you instructor helped you with in quantification theory?
From what you have posted, the answer to that question is: VERY LITTLE.

11. Originally Posted by Plato
Well I doubt that first statement. Is your idea of international is limited?
What has you instructor helped you with in quantification theory?
From what you have posted, the answer to that question is: VERY LITTLE.
My dear Plato you are simply curious to see what the right proof to that problem is .

Why ,don't you know the answer, or you agree with clic clac's proof??

12. Mhh.. Even before studying logic, one of the first rules in "advanced" mathematics you may see is: if $P$ is a proposition, $\neg(\forall xP)\equiv\exists x\neg P$ and $\neg(\exists xP)\equiv \forall x\neg P,$ and that makes sense (I think most people will agree with that).
Moreover, as you wrote yourself:
O.k for the first step since there is the theorem in predicate calculus to account for that,i.e

..............~.......................................
So I don't see the problem, you "replaced" $\forall$ by $\exists$ ...

My dear Plato you are simply curious to see what the right proof to that problem is
The right proof?

As you seem to know about predicate calculus and have a great teacher, Plato's reaction is quite understandable!

13. Originally Posted by clic-clac
The only things I used are what I said:
You don't have to get rid of the quantification symbols, just replace them by the other one.

For instance, from second step to third:

$\exists\epsilon[\epsilon>0\wedge\neg(\exists k(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$

$\equiv\exists\epsilon[\epsilon>0\wedge\forall k\neg(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$

But $\neg(k\in \mathbb{N}\wedge\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))$ has the form $\neg(A\wedge B)$ so it's the same than $k\in \mathbb{N}\Rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon))$

we replace it in the whole formula:

$\equiv\exists\epsilon[\epsilon>0\wedge\forall k(k\in\mathbb{N}\rightarrow\neg(\forall n( n\geq k\rightarrow|a_n-a|\leq\epsilon)))]$
Originally Posted by archidi
But there is no law of logic that allow you to do that ,at least as far as i know.

If however you happen to know of any such law ,mention it so that we could see how that law is derived from the axioms of the predicate calculus

All i said is that as far as i know there are not any laws of logic that allow you to replace any part of a formula by another equivalent formula

Do you know any such laws?Just state them and lets discuss them

Does that mean that i fancy my self as a logician?? As plato said ??

Really i dont see the point of writing useless posts confusing the viewers at the same time.

I do apologize to the viewers for writing posts completely irrelevant to the main issue of the problem

14. All i said is that as far as i know there are not any laws of logic that allow you to replace any part of a formula by another equivalent formula
Since your first post did not mention you were asking about replacing a formula by an equivalent one, I couldn't see what part of the proof you did not agree with.

Of course we have the right to do that! The proof isn't so simple and I can't write a whole course. That can be done by induction, first proving it for atomic formulas (that doesn't demand a lot of efforts) and then checking the different cases.

15. Originally Posted by clic-clac
Since your first post did not mention you were asking about replacing a formula by an equivalent one, I couldn't see what part of the proof you did not agree with.

Of course we have the right to do that! The proof isn't so simple and I can't write a whole course. That can be done by induction, first proving it for atomic formulas (that doesn't demand a lot of efforts) and then checking the different cases.

The only rules for substitution are the following ones:

1) $\vdash r=s\rightarrow t_{r}=t_{s}$,where r,s, $t_{r},t_{s}$ are terms,r occurs in $t_{r}$ and $t_{s}$ is the result of replacing one or more specified occurences ( but not necessarily all occurrences) of r in $t_{r}$ by occurrences of s.

......for e.g if r=s =====> x+r= x+s..............................................

Where now $t_{r}$ is :x+r,and $t_{s}$ is x+s............................................... ................................................

2) $\vdash r=s\rightarrow (P_{r}\Longleftrightarrow P_{s})$,where $P_{r},P_{s}$ are formulas,r and s are terms,r occurs in $P_{r}$ and $P_{s}$ is the result of replacing one or more specified occurrences ( but not necessarily all occurrences) of r in $p_{r}$ by occurrences of s,provided that no variable that occurs in r or s is a bound variable of $P_{r}$.

And a simple example is;

...........r=s ,then x<r <====>x<s......................................... ................................................

Where now $P_{r}$ is x<r ,and $P_{s}$ is x<s.....................................

Both of the above theorems are proved by induction.

As you can see none of them can be used in your proof

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