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Preimages problem

  1. Nov 7, 2011 #1
    [PLAIN]http://img855.imageshack.us/img855/5949/metric.jpg [Broken]

    If [itex]X,Y[/itex] are sets and [itex]f:X\to Y[/itex] is a function with [itex]B\subset Y[/itex], the preimage is defined [itex]f^*(B) = \{x\in X : f(x) \in B\}[/itex].

    If [itex]d_X, d_Y[/itex] are metrics on [itex]X,Y[/itex], continuity of [itex]f[/itex] can be characterised as follows:

    The preimage of any open (resp. closed) set in [itex](Y,d_Y)[/itex] is open (resp. closed) in [itex](X,d_X)[/itex].

    Hence, for example if we define [itex]f_1 (x,y) = x-y[/itex] then [itex]f_1[/itex] is continuous and [itex]A_1 = f_1^*\left( (-\infty ,1] \right)[/itex]. Since [itex](-\infty , 1][/itex] is closed, [itex]A_1[/itex] is closed.

    Similarly for [itex]A_2[/itex] and [itex]A_3[/itex], but not sure about [itex]A_4[/itex]. Can I write it in a way that makes it more obvious/easier to work with the preimage?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 7, 2011 #2

    HallsofIvy

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    Re: Preimages

    Left f(x,y)= x- y, a continuous function. The set of all (x, y) such that [itex]x- y\le 1[/itex] is the preimage, by that function, of the set [itex]\{z| z\le 1\}[/itex]. Since that is a closed interval, it follows that A1 is a closed sert.
     
  4. Nov 7, 2011 #3
    Re: Preimages

    I know that ([itex]A_1[/itex] is the one I proved already!) - it's [itexA_4[/tex] that I can't see how to do...
     
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