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Open sets

  1. Jan 17, 2009 #1
    How do we prove that the following set is open??


    S= [tex]R^2[/tex]- {(x,y) : y=[tex]x^2[/tex], xεR}

    WHERE R is for real Nos
     
  2. jcsd
  3. Jan 17, 2009 #2

    HallsofIvy

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    How about using the definition of "open set", which, in a metric space, is that every point of the set is an interior point. Let (x,y) be a point in S which means it is a pair of real numbers such that y is NOT equal to [itex]x^2[/itex]. Then show that there exist a distance r such that all points within r of (x,y) are also in S.

    One way to do that is to find the shortest distance from (x,y) to the curve [itex]y= x^2[/itex] and take half that distance as r.

    Frankly, it is hard to see how you could be expected to do a problem like this if you could not see how to prove that the limit of a constant sequence is that constant.
     
  4. Jan 17, 2009 #3
    Thanx, but how do we calculate that distance d(x,y) =r??
     
  5. Jan 17, 2009 #4
    distance formula?
     
  6. Jan 18, 2009 #5

    HallsofIvy

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    As long as you know there is such a non-zero distance, you don't need to know what it is!
     
  7. Jan 27, 2009 #6
    From the definition of the open set we have:

    S is open iff for all xεS THERE exists r>0 such that Disc(x,r) [tex]\subseteq S[/tex]

    .........................................or(equivalently).............................................


    S is open iff for all xεS ,there exists r>0 such that, yεDisc(x,r) =====> yεS.


    Let now x=(k,l) ,y=(m,n) then yεDisc(x,r) <=====>[tex]\sqrt{(k-m)^2+(l-n)^2}[/tex]< r

    AND the above becomes:



    S is open iff [if (k,l)εS then there exists r>0 such that ,if [tex]\sqrt{(k-m)^2+(l-n)^2}[/tex]< r then (m,n)εS].


    But (k,l)εS <====> k< [tex]\l^2[/tex], (m,n)εS <====> m< [tex]\ n^2[/tex]. And if w=(o,p) is a point on the curve THEN the distance between y and w is :

    .............[tex]\sqrt{(m-o)^2 + (n-p)^2}[/tex].............................................

    AND if we choose r< [tex]\sqrt{(m-o)^2 + (n-p)^2}[/tex]....the above becomes:



    S is open iff .

    ...................if....... k< [tex]\l^2[/tex] and o=p^2 and [tex]\sqrt{(k-m)^2+(l-n)^2}[/tex]< [tex]\sqrt{(m-o)^2 + (n-p)^2}[/tex].......... then m< [tex]\ n^2[/tex]

    Can anybody curry on from here???

    Also note k< [tex]\l^2[/tex] is one case another is k> [tex]\l^2[/tex]
     
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