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Seemingly simple problem

  1. May 8, 2008 #1
    Let [tex]R[/tex] be the set of real numbers and let [tex] <, \leq [/tex] have their usual meanings.

    Show that [tex] \{ x \in R : 0 \leq x \leq 3\} \cap \{ x \in R : -1 < x < 1\} = \{ x \in R : 0 \leq x < 1\} [/tex]

    This seems very straight forward, but I am having trouble with wording and figuring out what is sufficient enough to conclude the proof.

    So here is what I would argue.
    By the definition of the intersection of two sets we have
    [tex] \{ x \in R : 0 \leq x \leq 3\} \cap \{ x \in R : -1 < x < 1\} = \{ x \in R : 0 \leq x \leq 3 and -1 < x < 1 \} [/tex]

    Here is where I am not sure about the language that I need to use...

    Can I just say something like:

    The only way both conditions can be true is if [tex]0 \leq x < 1[/tex]

    Therefore we have [tex]\{ x \in R : 0 \leq x \leq 3 and -1 < x < 1 \} = \{ x \in R : 0 \leq x < 1\}[/tex]

    Or should I not treat this as obvious???

    Or is there a better way to approach this all together?

    My definition of intersection is as follows:

    [tex] X \cap Y = \{x : x\in X and x\in Y\} [/tex]

    Up to this point there is no formal treatment of either open and closed intervals, so I am not sure how informal or obvious I should assume them to be.
     
  2. jcsd
  3. May 8, 2008 #2
    I don't know how if this notion fits your formality target, but I think the idea of open and closed intervals is already embedded in the [tex]<[/tex] and [tex]\leq[/tex] operators. When a mathematical proof uses a statement like "have their usual meaning", I'd guess a little informality would be tolerated.
     
  4. May 9, 2008 #3

    tiny-tim

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    Hi Diffy! :smile:

    I suspect it's a logic test, and they want you to say something like …

    if x > a and x > b, then x > sup{a,b}. :smile:
     
  5. May 9, 2008 #4
    I guess you can break it up further to show why exactly it's like that.

    {x in R | 0 <= x <= 3} int {x in R | -1 < x < 1 }

    = [{x in R | x >= 0) int {x in R | x <= 3}] int [{x in R | x > -1} int {x in R | x < 1}] <-- Break it down to separate intersections of sets

    = [{x in R | x >= 0) int {x in R | x > -1}] int [{x in R | x <= 3} int {x in R | x < 1}] <-- Rearrange the sets since we're dealing with all set intersections

    = {x in R | x >= 0) int {x in R | x < 1} <--simplify to satisfy set conditions when intersected

    = {x in R | 0 <= x < 1} <-- simplify further to satisfy set conditions when intersected
     
  6. May 9, 2008 #5
    Hi everyone,

    Tiny Tim, that seems to make a lot of sense. Even though the author never uses Sup or Inf, I will go ahead an proceed that way.

    I am still a bit confused though on having "less than" and "Less than or equal to"

    So I will say since x < 1 and [tex] x \leq 3[/tex], then we must have x ? inf(1, 3)

    where the question mark is < or [tex]\leq[/tex]?

    BryanP, I think you are saying the same thing that TinyTim is, so thanks for the help, I think breaking it up into that many pieces is missing the main point of determining how we know that the intersections of these intervals are what they are.

    BobPi,
    I would have rounded 3141592 up to 3141593.

    Anyways, you are right, I think I can be very informal, but I was just missing the piece on how we know if x<1 and x < 3 that x must be less than 1, using least upper bounds and greatest lower bounds is the key I think.
     
    Last edited: May 9, 2008
  7. May 9, 2008 #6

    tiny-tim

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    Hi Diffy! :smile:
    For the "≤"? Just copy-and-paste what someone else has done!
    ah … you need something like lowest-common-multiple in fractions.

    Try rewriting one of the sets {x < 1} and {x ≤ 3} so that it matches the other …

    {x < 1} = {x ≤ 1} ∩ {…?} :smile:
    But Bob is modest, and a model of understatement! :blushing:
     
  8. May 9, 2008 #7

    Nah, I forgot a / in my tex tag, so the whole post looked very strange.

    Alrighty,
    {x < 1} = {x ≤ 1} ∩ {x < 1}?

    or

    {x < 1} = {x ≤ 1} ∩ {x ≤ ?}


    I don't see where you are going.. Sorry I'm a little slow and I have trouble these details.


    lol I know, I was just teasing, all in fun.
     
  9. May 9, 2008 #8

    tiny-tim

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    Hi Diffy! :smile:

    How about
    {x < 1} = {x ≤ 1} ∩ {x ≠ 1} ? :smile:
     
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