- #1
rukawakaede
- 59
- 0
As title, is this true ?
(x,x]={x}?
(x,x]={x}?
rukawakaede said:As title, is this true ?
(x,x]={x}?
rukawakaede said:As title, is this true ?
(x,x]={x}?
tiny-tim said:hi rukawakaede! no, (x,x] is empty
tiny-tim said:yes
rukawakaede said:how about
[tex]\bigcup_{n\in\mathbb{N}}(x-\frac{1}{n},x]=\{x\}[/tex]
?
Fredrik said: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 [itex]\mathbb Z^+[/itex] for the positive integers.
SW VandeCarr said:If you're going to write it that way, shouldn't that be [tex] n\in\mathbb R[/tex]?
rukawakaede said:sorry for my lack of clarity.
n is in [tex]\mathbb{N}[/tex]
as I use [tex]\mathbb{N}[/tex] as a countable index set.
sorry i didn't mention [tex]x\in\mathbb{R}[/tex], but I think it is clear from context.
He's just asking ifSW VandeCarr said: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.
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 said: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 said: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 said: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}.
tiny-tim said: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:
SW VandeCarr said: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"
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).SW VandeCarr said:The interval (b,a] is open to b (the interval doesn't include b) but closed in 'a' (the interval includes a.)
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 said: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 said: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
Fredrik said: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]=∅.
A well-ordered set is a mathematical concept that describes a set in which every non-empty subset has a least element. This means that every element in the set has a clear and defined order, and there are no "gaps" or missing elements in the set.
In a totally ordered set, every pair of elements can be compared and there is a clear order between them. However, in a well-ordered set, every non-empty subset has a least element, not just every pair of elements. This means that a well-ordered set has a more strict and defined structure compared to a totally ordered set.
Yes, (a,b] is a well-ordered set if a and b are real numbers and a < b. This is because every non-empty subset of (a,b] has a least element. For example, the subset (a,b] itself has a least element of a, and any other subset within (a,b] will also have a least element.
Yes, a set can be both well-ordered and infinite. One example is the set of natural numbers (1, 2, 3, ...), which is both infinite and well-ordered since every non-empty subset has a least element (e.g. the subset {1,2,3} has a least element of 1).
Well-ordered sets are used in various branches of mathematics, including set theory, real analysis, and topology. They provide a foundational concept for understanding the structure of sets and can be used to prove theorems and make logical deductions in these fields.