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Show that the half open interval is a topology

  1. Dec 6, 2013 #1
    1. The problem statement, all variables and given/known data
    We are given ##H## = {##O | \forall x, \exists a,b \in R## s.t ##x \in [a,b] \subseteqq O##}##\bigcup {\oslash}##

    and are asked to show that it is a topology on R

    2. Relevant equations
    Definition of a topological space


    3. The attempt at a solution

    I am trying to convince myself that R is in H. I do not see it. This is, of course, the first property of a topological space.
     
  2. jcsd
  3. Dec 6, 2013 #2
    So take ##x\in \mathbb{R}##, you must find ##a,b## such that ##x\in [a,b]\subseteq \mathbb{R}##. Any idea of what the ##a## and ##b## are gonna be?? (Hint: any ##a## and ##b## are good as long as ##a\leq b##)
     
  4. Dec 7, 2013 #3

    Dick

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    Since for any x, x is an element of [x,x], the way you stated the first part of your union, the only subset O of R that will satisfy that is R itself. There is something wrong with your problem statement. That doesn't look like a half-open topology to me.
     
  5. Dec 7, 2013 #4
    OK so now I am trying to show that a finite intersection is in H. I am still really confused on what this set actually is. I would appreciate if anyone could explain to me how this set is/works.
     
  6. Dec 7, 2013 #5

    vela

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    Why does the thread title refer to the half-open interval topology? I don't see any half-open intervals in your definition.

    You do realize that for any set A, its union with the empty set is itself, right? So you might as well have written
    $$H = \{ O: \forall x, \exists a,b \in \mathbb{R} \text{ such that } x \in [a,b] \subset O \}.$$ What you probably meant was
    $$H = \{ O: \forall x, \exists a,b \in \mathbb{R} \text{ such that } x \in [a,b] \subset O \} \cup \{\emptyset\},$$ so that the empty set is an element of H.

    Finally, what about the point Dick brought up? The only set that satisfies your definition is ##\mathbb{R}## since ##x## can be anything, so this topology is the trivial topology ##\{ \emptyset, \mathbb{R}\}##, not the half-open interval topology.

    You really need to get the definition of H down before you go about trying to prove anything with it.
     
  7. Dec 7, 2013 #6

    Dick

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    The half-open interval topology or lower limit topology is usually defined as the topology generated by the basis sets [a,b). Your set definition doesn't resemble that in the slightest.
     
  8. Dec 8, 2013 #7
    My mistake people. There was suppose to be an open bracket at b. [a,b)
     
  9. Dec 8, 2013 #8

    Dick

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    Ok, so your topology consists of the set of all sets O such for every point x in O, there is an interval [a,b) that contains the point x and is a subset of O. Does that make it clearer what your topology is?
     
    Last edited: Dec 8, 2013
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