Is (a,b] a well-ordered set?

[rukawakaede]

As title, is this true ?

(x,x]={x}?

 

[SW VandeCarr]

As title, is this true ?

(x,x]={x}?

I would not know how to interpret that. It seems if you want to indicate a half-open interval you would write (b,a]. The variable x could then be defined as $$b> x \geq a$$.

 

[tiny-tim]

hi rukawakaede!

As title, is this true ?

(x,x]={x}?

no, (x,x] is empty

 

[rukawakaede]

hi rukawakaede! no, (x,x] is empty

how about

$$\bigcap_{n\in\mathbb{N}}(x-\frac{1}{n},x]=\{x\}$$

?sorry, i made a typo just now. now corrected.

 

[tiny-tim]

yes

 

[rukawakaede]

yes

Thank you!

 

[Fredrik]

I guess (x,x] would have to be defined as the set of all real numbers t such that x<t≤x. There is no such t (if one of the inequalities hold, then the other doesn't), so (x,x] must be the empty set.

 

[SW VandeCarr]

how about

$$\bigcup_{n\in\mathbb{N}}(x-\frac{1}{n},x]=\{x\}$$

?

If you're going to write it that way, shouldn't that be $$a\in\mathbb R$$?

 

[Fredrik]

I think he meant to make that an intersection, not a union. As it stands, the left-hand side is =(x-1,x], assuming that we don't count 0 as a natural number. I prefer to include 0 in the natural numbers, and to use $\mathbb Z^+$ for the positive integers.

 

[rukawakaede]

I think he meant to make that an intersection, not a union. As it stands, the left-hand side is =(x-1,x], assuming that we don't count 0 as a natural number. I prefer to include 0 in the natural numbers, and to use $\mathbb Z^+$ for the positive integers.

i made a typo. thanks for correction.
:)

and also $$\mathbb{N}=\{1,2,3,\cdots\}$$

I don't know why I tend to write natural number including 0 in this way: $$\mathbb{N}_0=\{0,1,2,3,\cdots\}$$

 

[rukawakaede]

If you're going to write it that way, shouldn't that be $$n\in\mathbb R$$?

sorry for my lack of clarity.
n is in $$\mathbb{N}$$
as I use $$\mathbb{N}$$ as a countable index set.

sorry i didn't mention $$x\in\mathbb{R}$$, but I think it is clear from context.

 

[SW VandeCarr]

sorry for my lack of clarity.
n is in $$\mathbb{N}$$
as I use $$\mathbb{N}$$ as a countable index set.

sorry i didn't mention $$x\in\mathbb{R}$$, but I think it is clear from context.

I thought any open or half-open interval implied an interval on the real number line. I'm not sure where countable sets are relevant. No one else has raised this, so I must be missing something. Perhaps you or someone else could explain. If you have (x,x] then you have x=a and x<b so x=a. This is not an empty set because {x=a} contains the point a although (x,x] is an interval of zero measure..

 

[Fredrik]

I thought any open or half-open interval implied an interval on the real number line. I'm not sure where countable sets are relevant. No one else has raised this, so I must be missing something.

He's just asking if

$$\bigcap_{n\in Z^+}\big(x-\frac{1}{n},x\big]=\{x\}$$

and it is, since x is a member of (x-1/n,x] for all n, and no other real number is. I'm not sure what bothers you about this. (He wrote \bigcup instead of \bigcap by accident, and I assume that his $\mathbb N$ is equal to what I call $\mathbb Z^+$, i.e. {1,2,3,...}). He could of course also have asked if

$$\bigcap_{r\in\mathbb R,\, r>0}\big(x-\frac{1}{r},x\big]=\{x\},$$

but that would just have been a slightly different way to make the same point.

If you have (x,x] then you have x=a and x<b so x=a. This is not an empty set because {x=a} contains the point a although (x,x] is an interval of zero measure..

I don't quite understand what you're saying here, but what I said in #7 explains why (x,x] is empty for all x.

 

[SW VandeCarr]

I don't quite understand what you're saying here, but what I said in #7 explains why (x,x] is empty for all x.

There's a difference between an interval of zero measure and the empty set. The interval (b,a] is open to b (the interval doesn't include b) but closed in 'a' (the interval includes a.) However 'a' is a point and if the interval contains only 'a', it is said to be of zero measure. Now I'm not sure exactly what (x,x] means, but it must at least mean that x takes the value 'a' if not any other values less than b. If so, the set {x} is not empty because {a}=(x}.

I also don't understand why you were talking about countable sets. It seems to me the notation (x,x] implies an interval on the real number line unless otherwise specified.

 

[tiny-tim]

Now I'm not sure exactly what (x,x] means, but it must at least mean that x takes the value 'a' if not any other values less than b. If so, the set {x} is not empty because {a}=(x}.

I'm with Fredrik on this …

(x,x] is open to the left and closed to the right: if it's open to the left, it can't contain x

alternatively, treating the reals merely as an ordered set without a topology, (x,x] contains elements strictly greater than x and less than or equal to x: it can't contain x

 

[SW VandeCarr]

I'm with Fredrik on this …

(x,x] is open to the left and closed to the right: if it's open to the left, it can't contain x

alternatively, treating the reals merely as an ordered set without a topology, (x,x] contains elements strictly greater than x and less than or equal to x: it can't contain x :win:k:

Yo no entiendo. Please explain. If the interval (b,a] contains 'a'; why can't x=a? Moreover in the next paragraph you say:

"alternatively, treating the reals merely as an ordered set without a topology, (x,x] contains elements strictly greater than x and less than or equal to x: it can't contain x"

Besides, why do we want to assume the real number line doesn't have a topology? It does. It is totally ordered (by the Axiom of Choice I believe) and is also a metric space, both features defining a topology.

 

[tiny-tim]

Yo no entiendo. Please explain. If the interval (b,a] contains 'a'; why can't x=a?

why should it?

the interval (3,2] doesn't contain 2, so why should (2,2] ?

Moreover in the next paragraph you say:

"alternatively, treating the reals merely as an ordered set without a topology, (x,x] contains elements strictly greater than x and less than or equal to x: it can't contain x"

yes, elements "strictly greater than x and less than or equal to x" don't include x

 

[Fredrik]

The interval (b,a] is open to b (the interval doesn't include b) but closed in 'a' (the interval includes a.)

If you define (b,a] as "the interval from b to a, with b not included and a included", then (x,x] is neither empty nor non-empty, it's just nonsense. (A set can't both contain x and not contain x). If you define (b,a] as "the set of all t such that b<t≤a", then (x,x] is empty. (I prefer the latter definition because it makes sense for all a and b, and doesn't require us to have defined "interval" in advance).

I also don't understand why you were talking about countable sets. It seems to me the notation (x,x] implies an interval on the real number line unless otherwise specified.

I don't understand your concern at all. n is an index that labels the sets in the family of sets that we're taking the intersection of. If the definitions of the sets in the family hadn't involved some algebraic operations, we could have used any set as the index set. In this particular case, where the sets are (x-1/n,x], the index set can be any subset of the real numbers that doesn't include 0.

 

[SW VandeCarr]

why should it?

the interval (3,2] doesn't contain 2, so why should (2,2] ?yes, elements "strictly greater than x and less than or equal to x" don't include x

I'm sorry. Why do you say that (3,2] doesn't include 2? As a half closed interval doesn't this notation correspond to $$3> x \geq 2$$? Please explain why you are saying 2 is not included in the interval when every reference I've checked says it is. I'll just post one.

http://mathworld.wolfram.com/Half-ClosedInterval.html

 

[Fredrik]

No, (3,2] is the set of all real numbers x such that 3<x≤2. There are no such real numbers. Therefore, (3,2]=∅.

 

[SW VandeCarr]

No, (3,2] is the set of all real numbers x such that 3<x≤2. There are no such real numbers. Therefore, (3,2]=∅.

OK. But that's not the inequality I wrote in my previous posts. However for (a,b] I do see your point. If I write $$a < x \leq b$$ then the corresponding set has no least element and therefore cannot be well ordered. Thanks.