How do these statements differ?

[dipole]

More than once I have been marked wrong for this, and it's beginning to annoy me...

How is the statement,

|x| < 1

different from the statement,

|x| \leq s < 1 for some s ?

This comes up when talking about the radius of convergence for some series.

 

[lostcauses10x]

One statement is a partial of the other.

 

[dipole]

If that's supposed to be a joke or something, I really don't get it. I'm asking how they logically differ and what difference it makes about the statement of convergence.

 

[lostcauses10x]

dipole
What is the relation to s in the first and only the first statement?
It does not exist.
Therefore it is a partial statement of the second one. It usually will be marked wrong.

 

[pwsnafu]

Assuming we are talking about the reals
$|x| < 1$ means the open interval $(-1,1)$
$|x| \leq s < 1$ means the closed interval $[-s,s]$ for some $0<s<1$.

Hint: consider $\frac{1+s}{2}$

 

[Office_Shredder]

Why don't you tell us the context in which one is supposed to matter and the other doesn't. Typically it is relevant when you are talking about uniform convergence: the series
\sum_{n\geq 0} x^n
converges in (-1,1) and converges uniformly in [-s,s] for any s < 1. So when doing proofs about integrating and differentiating the power series that require uniform convergence to hold you have to restrict yourself to [-s,s].

 

[1MileCrash]

I see no difference between the two either..

 

[R136a1]

Why don't you tell us the context in which one is supposed to matter and the other doesn't. Typically it is relevant when you are talking about uniform convergence: the series
\sum_{n\geq 0} x^n
converges in (-1,1) and converges uniformly in [-s,s] for any s < 1. So when doing proofs about integrating and differentiating the power series that require uniform convergence to hold you have to restrict yourself to [-s,s].

I agree, you should show us the correct context.

 

[HallsofIvy]

"|x|< 1" refers to any number satisfying -1< x< 1. The set of such numbers includes numbers "arbitrarily" close to 1.

"|x|\le s for some s< 1" posits some specific bound on x less than 1. The set of such numbers does NOT include number "arbitrarily" close to 1.

 

[AlephZero]

The reason the two statements are different is: There is no real number that is the less than 1 but the "nearest" number to 1.

A statement like yours would be correct if you were talking about integers. For example if N is an integer and $|N| < 10$, then there is an integer S such that $|N| <= S < 10$, and of course S = 9.

But the same logic doesn't work for real numbers, or rationals, because you can always find another number in between ANY pair of real or rational numbers.

Whatever number you choose for s, I can choose another number closer to 1 - for example (1+s)/2. So the number you call "some s" doesn't exist.

This logical error is similar to the wrong idea that the number written as "0.999999..." in decimal notation is somehow different from the number "1".

 

[1MileCrash]

OK, so is the issue that we can't represent an open interval as a closed interval with different endpoints, because these endpoints would have to be, essentially, hyperreal?

 

[R136a1]

OK, so is the issue that we can't represent an open interval as a closed interval with different endpoints, because these endpoints would have to be, essentially, hyperreal?

An open interval has no endpoints, not even hyperreal ones.

 

[1MileCrash]

An open interval has no endpoints, not even hyperreal ones.

Then replace my use of "endpoints" with "numbers that are next to the little parenthesis or bracket things in the notation" if you'd prefer.

I don't know what you'd call x and y in the interval (x,y) besides endpoints. They aren't included in the interval, but I've never heard anyone hesitate to call them endpoints.

 

[R136a1]

Then replace my use of "endpoints" with "numbers that are next to the little parenthesis or bracket things in the notation" if you'd prefer.

I don't know what you'd call x and y in the interval (x,y) besides endpoints. They aren't included in the interval, but I've never heard anyone hesitate to call them endpoints.

I guess endpoint is a suitable name, but you seemed to imply that they belong to the interval, which they don't. You could always use suprema or infima.

 

[1MileCrash]

I guess endpoint is a suitable name, but you seemed to imply that they belong to the interval, which they don't. You could always use suprema or infima.

What I'm asking / saying is that from the replies above, if the OPs error is similar/equivalent to the idea that I cannot represent the interval:

(x, y)

As a closed interval

[a, b]

By defining new numbers a and b, because in order for the intervals to be equivalent, a and b would have to be defined in a way that made them not real numbers.

 

[Stephen Tashi]

More than once I have been marked wrong for this, and it's beginning to annoy me...

How is the statement,

|x| &lt; 1

different from the statement,

|x| \leq s &lt; 1 for some s ?

Technically those are not statements. To be a statement, you would have to give a complete context for those notations. For example, "There exists a number x such that |x| < 1". A statement must be a claim that is True or False.

Interpreting |x| < 1 to mean "the set of numbers x such that |x| < 1" doesn't produce a statement. It produces a noun.

Perhaps you mean to ask whether the two notations are notations for the same set of numbers. Interpreting |x| <1 as a specification of set defines a particular set. The second notation does not define a particular set unless there is more context that defines a particular s. Perhaps the context you are using is "There exists a number s such that s < 1 and the series converges for all numbers x such that |x| < s". That statement does not comment on whether the series converges for a number y such that s < y < 1.