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Math Help - Usual topology

  1. #1
    MHF Contributor Amer's Avatar
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    Jordan
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    Usual topology

    My professor gave us a question in exam

    Let A=[-2,2] and consider the subspace (A,T_A) of the space (R,T_u) .Take

    B=\{x : \mid x \mid <1\}-(\{\frac{1}{n} : n\in N \})

    find a T_u open subset W\subseteq R such that
    B=A\cap W . Conclude that B is T_A open

    I answered W=B since B is open in T_u and B\subseteq A so A\cap W = A\cap B=B

    but after he finished the correction of the exam and he want to gave us it he said that the question is deleted since he forgot to set a side {0} from B so B present like this

    B=\{x : \mid x \mid <1\}-(\{\frac{1}{n} : n\in N \}\cup \{0\})

    and he said that B in the first form is not T_u open !! and I ask him why?? he answered me but I did not got it and I shamed to ask him again .

    so can you explain why B in the first form is not open although it is an arbitrary union of open sets in T_u and the arbitrary unoin of open set is open .

    and we can write B like this B=(-1,\frac{1}{2})\cup (\frac{1}{2},\frac{1}{3})\cup (\frac{1}{3},\frac{1}{4})\cup (\frac{1}{4},\frac{1}{5})\cup.....

    Thanks
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  2. #2
    Super Member Gamma's Avatar
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    Iowa City, IA
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    Quote Originally Posted by Amer View Post

    and we can write B like this B=(-1,\frac{1}{2})\cup (\frac{1}{2},\frac{1}{3})\cup (\frac{1}{3},\frac{1}{4})\cup (\frac{1}{4},\frac{1}{5})\cup.....

    Thanks
    See I think part of the propblem is your fractions are screwed up, that first set in the union is going to contain everything else in your union, so all that is unnecessary, and i think we can all agree B \not = (-1, \frac{1}{2}). Recall  \frac{1}{n}\rightarrow 0.
    I see what you are trying to do though and see the confusion.

    By definition, a set B is open iff for every x in B, there is an open set containing x and contained in B.

    But consider that point 0 that is in B. There is no open interval (these are the basis elements for the real line in the usual topology) which contains 0 and does not intersect the sequence  \{\frac{1}{n}\} (in fact there are an infinite number of points of this intersection for every open set containing 0 by definition of  \frac{1}{n}\rightarrow 0).

    If instead 0 were removed like the teachers correction states (call it B') what you were trying to do is spot on.
    B'=(-1,0)\bigcup_{n=1}^\infty(\frac{1}{n+1},\frac{1}{n}  )
    but the way it stands as he originally stated, 0 is in B, but not in B'.
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Gamma View Post
    See I think part of the propblem is your fractions are screwed up, that first set in the union is going to contain everything else in your union, so all that is unnecessary, and i think we can all agree B \not = (-1, \frac{1}{2}). Recall  \frac{1}{n}\rightarrow 0.
    I see what you are trying to do though and see the confusion.

    By definition, a set B is open iff for every x in B, there is an open set containing x and contained in B.

    But consider that point 0 that is in B. There is no open interval (these are the basis elements for the real line in the usual topology) which contains 0 and does not intersect the sequence  \{\frac{1}{n}\} (in fact there are an infinite number of points of this intersection for every open set containing 0 by definition of  \frac{1}{n}\rightarrow 0).

    If instead 0 were removed like the teachers correction states (call it B') what you were trying to do is spot on.
    B'=(-1,0)\bigcup_{n=1}^\infty(\frac{1}{n+1},\frac{1}{n}  )
    but the way it stands as he originally stated, 0 is in B, but not in B'.
    thanks very much I understand it .
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