Question

\[a\le b+\epsilon\] for all epsilon greater than zero
Then \[a\le b\]

Am I stating it wrong?

Answer 1

may or may not be correct, depends on the value of epsilon

Answer 2

is this a Cauchy thing?

Answer 3

Or consider the following,
\[a\le b+\epsilon \] for all epsilon greater than zero.
Then \[a<b\]

Is this statement correct, or the previous one?

Answer 4

is that supposed to be
if \[a\le b+\epsilon,\epsilon>0\]then\[a<b\]??

Answer 5

that is what I am asking..

Answer 6

um... yeah, looks like it.

Answer 7

So which statement is correct ?
Should it be \[a<b\] or \[a\le b\]

Answer 8

Your first statement of the problem is correct

Answer 9

a≤b+ϵ for all epsilon > 0 implies a≤b

Answer 10

if \[\epsilon>0\]then it should be \[a<b\]if we allowed\[a=b\]adding positive epsilon would make\[a>b\]then we would have\[a<b+\epsilon\]which is not what you wanted

Answer 11

scratch the a>b, typo

Answer 12

Well, I have a kind of feeling, its as follows.
Suppose a is lesser than b for all epsilon added to b, then its obvious that a is lesser than b
But if a is equal to b for some small epsilon, then a must become less than b when epsilon is removed

Answer 13

I've told you the answer. Prove it rigorously.

And in a way it's obvious that it can't that the conclusion is a < b. For example if a = b. Then
\[ a≤b + \epsilon \]
for all \( \epslion > 0 \).

Answer 14

Ok thanks. Got you...

Answer 15

James do you agree with this, that no matter what a and b I take, the relation a=b+ epsilon is never going to happen

Answer 16

It can't, because if it did for some particular \( \epslion' \), then
\[ a > b + \epsilon'/2 \] which violates the hypothesis.

Answer 17

So do you think that the theorem would hold even if it were stated as follows
a<b+ϵ for all epsilon > 0 implies a≤b
Am I right?

Answer 18

do you think that the theorem would hold even if it were stated as follows
a<b+ϵ for all epsilon > 0 implies a≤b
Am I right?

Answer 19

That's correct.

Answer 20

Ok then what is the point of that equality?

Answer 21

The thing that is given is
\[a\le b+\epsilon\]

Answer 22

What does that mean?
Either \[a< b+\epsilon\]
or
\[a= b+\epsilon\]
implies that \[a\le b\]
Am I right?

Answer 23

There are four possible theorems here

1. \[ (\forall \epsilon > 0) \ a \leq b+\epsilon \implies a \leq b \]
2. \[ (\forall \epsilon > 0) \ a \leq b+\epsilon \implies a < b \]
3. \[ (\forall \epsilon > 0) \ a < b+\epsilon \implies a \leq b \]
4. \[ (\forall \epsilon > 0) \ a < b+\epsilon \implies a < b \]

Which of these is true?

Answer 24

I think the first and the third, am I right?

Answer 25

We're seen why 2 is false. Why is 4 false?

Answer 26

Four is not false, but four is just a particular case of 3, right?

Answer 27

Well my question is the same still
Does the theorem means (implicitly )
Either
a<b+ϵ

or
a=b+ϵ

implies that
a≤b

Am I right?

Answer 28

\[ (\forall \epsilon > 0) \ a \leq b+\epsilon \implies ( (\forall \epsilon > 0) \ a < b+\epsilon) \ or \ ((\forall \epsilon > 0) \ a = b+\epsilon) \]
and
\[ (\forall \epsilon > 0) \ a \leq b+\epsilon \implies a \leq b \]

But that doesn't necessarily mean
\[ ( (\forall \epsilon > 0) \ a < b+\epsilon) \ or \ ((\forall \epsilon > 0) \ a = b+\epsilon) \implies a \leq b \]

In other words,
(A => B) and (A => C) doesn't necessarily mean (B => C).

But in this case
\[ ( (\forall \epsilon > 0) \ a < b+\epsilon) \ or \ ((\forall \epsilon > 0) \ a = b+\epsilon) \implies ( (\forall \epsilon > 0) \ a < b+\epsilon) \]
because the second statement is false. And indeed
\[ ( (\forall \epsilon > 0) \ a < b+\epsilon) \implies a < b \implies a \leq b \]

Answer 29

I will not be wasting your time keeping you waiting on this. At present my brain is trying to understand your last post. I will let you know once the processing is over.

Answer 30

But before you leave, I am once again trying to present my question by saying, what is the need of "less than equal to", when the theorem is OK for just

a<b+ϵ for all epsilon > 0 implies a≤b

Answer 31

Actually there's a mistake in my last line
\[ (\forall \epsilon>0) a < b+ \epsilon) \implies a \leq b \]
but it doesn't imply a < b.

Therefore in fact theorem 4 above is false. For example if a = b, then
\[ a < b + \epsilon \ \forall \epsilon > 0 \]

Answer 32

A while ago you agreed that no matter what value of a and b I take, I will never get a=b+\epsilon
Then what is the point of keeping that equality in the theorem

Answer 33

What we've shown is
\[ (\forall \epsilon>0) a \leq b + \epsilon \implies (\forall \epsilon>0) a < b + \epsilon \]
So the equality is irrelevant.

Answer 34

Yes, "equality is irrelevant.", then why keep it?

Answer 35

because sometimes when you come to use these results in other more important theorems, you'll have an inequality with
\[ \leq \] versus with \[ < . \] It's a useful technical thing to know that it doesn't matter; that in both cases you can draw the same conclusion.

All of these results are just technical lemmas that are used for more important results. They're also a good training ground for gaining familiarity for these sorts of issues.

Answer 36

Ok then, I will need this later, and now I am satisfied.
Now if you don't mind, may I ask, did you find anyone else, before, bothering himself with such a question for such a long time. I mean do you think I was a slow thinker

Answer 37

The step up in thinking required in introductory analysis to begin manipulating with epsilon-type proofs (like this one; like epsilon-N for sequences and series; like epsilon-delta for continuous functions) is a big step for just about everyone I know.

I wouldn't feel bad about it. But do work lots and lots of problems with it so you gain fluency.

Just as in HS you didn't "get" quadratic equations the first day you were shown them, but after being forced to solve 100+ of them, you can now do them easily. This is intellectually analogous.

Answer 38

James you don't have the slightest idea how helpful you are.
Thanks once again.

Answer 39

Um thanks. All cash donations gratefully received. ;-)

Answer 40

How to send?

Answer 41

joke!

Answer 42

til later

Answer 43

lol

Answer 44

Don't consider this a joke, I am serious, I need your comment on my following statement.

\[( (\forall \epsilon > 0) \ a < b+\epsilon) \ or \ ((\forall \epsilon > 0) \ a = b+\epsilon)\ or\ ((\forall \epsilon > 0) \ a = banana)\ or \ ((\forall \epsilon > 0) \ a = grapes) \]
\[\implies ( (\forall \epsilon > 0) \ a < b+\epsilon)\implies a\le b\]

Answer 45

Your equation is chopped off. But the basic principle is

A or B or C or D or E => A
if B, C, D and E are false.

Answer 46

Here it is
(∀ϵ>0) a<b+ϵ) or ((∀ϵ>0) a=b+ϵ) or ((∀ϵ>0) a=banana) or ((∀ϵ>0) a=grapes)

=>((∀ϵ>0) a<b+ϵ)=>a≤b

Answer 47

yes

Answer 48

because a=banana and a=grapes is false for any value of epsilon

Answer 49

Finally done! Now I am going to write it down somewhere.. Thanks again

Answer 50

actually, there is one quibble here...

Answer 51

\[ (\forall \epsilon >0) \ a \leq b + \epsilon \] means that for each such epsilon
\[ a < b + \epsilon \]
or
\[ a = b + \epsilon \]

The more logically coherent way to show then that equality is never attained is to imitate the argument we wrote above:

If there any particular \( \epsilon > 0 \), call it \( \epsilon' \) such that
\[ a = b + \epsilon' \]
then it must be the case that
\[ a > b + \epsilon'/2 \]
which contradicts the hypothesis.

Hence \[ (\forall \epsilon >0) \ a \leq b + \epsilon \implies (\forall \epsilon >0) \ a < b + \epsilon \]

Answer 52

Yes, I understand that, but where is the quibble. Is it that, I can't actually show that apple and grape is not possible?

Answer 53

....just like you showed equality is not possible

Answer 54

No, actually, sigh, I'm embarrassed and annoyed that what I wrote down above isn't correct. One sec ... I'll write it out.

Answer 55

Consider the following:
\[ (\forall n \in \mathbb{N}) ( (n \hbox{ is even}) \ or \ (n \hbox{ is odd})) \]
That doesn't imply
\[ ((\forall n \in \mathbb{N}) (n \hbox{ is even})) \ or \ ((\forall n \in \mathbb{N}) (n \hbox{ is odd})) \]

so far so good?

Answer 56

OK...

Answer 57

Likewise
\[ (\forall \epsilon >0)\ a \leq b + \epsilon \]
doesn't imply
\[ ( (\forall \epsilon >0)\ a < b + \epsilon ) \ or \ ( (\forall \epsilon >0)\ a = b + \epsilon ) \]

If we want to show (and we do!) that
\[ (\forall \epsilon >0)\ a \leq b + \epsilon \implies (\forall \epsilon >0)\ a < b + \epsilon \]
use the new argument I've just written above. Not the old one, which is wrong.

Answer 58

The general logical principle here is this:
\[ (x \in S) (P(x) \ or \ Q(x)) \] doesn't imply
\[ ((x \in S) P(x) ) \ or \ ((x \in S) Q(x)) \]

Answer 59

Yes, I am getting you. Could you please copy paste the part that you wrote wrong previously

Answer 60

This step is wrong:

Answer 61

OK..

Answer 62

sorry bout that. (and see? even old(er) dogs make errors.)

Answer 63

LOL

Answer 64

So finally may I say
(∀ϵ>0) (a<b+ϵ or a=b+ϵ or a=banana or a=grapes)

=>((∀ϵ>0) a<b+ϵ)=>a≤b

Answer 65

Yes. But reasons in each case are different

It can't be the case that
a=b+ϵ
for any epsilon by the argument above

It's not the case that
a=banana
or
a=grapes
because a is a number, not a fruit.

The general logical principle here then is
\[ [(\forall x \in S) (P(x) \lor Q(x))] \land [(\forall x \in S) \lnot Q(x)] \implies (x \in S) P(x) \]
where \( \lor \) is or and \( \lnot \) is not