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A sequence or a single point?

  1. Apr 4, 2010 #1
    Suppose we define a set A to consist of all sequences x=<x_i> of real numbers, for which some condition holds, define a metric on it, and show that it generates a topology T on A.

    What i am a little unclear about is when we try to show if (A,T) is Hausdorff or not, do we pick now two points(single real numbers) x,y from A, and show that there are(are not) neighgorhoods U,V or x and y respectively that are disjoint, or do we pick sequences x and y of real numbers, instead?

    My intuition says they should be sequences, but not quite sure about it.

    Thanks for your help?
    Last edited: Apr 4, 2010
  2. jcsd
  3. Apr 4, 2010 #2
    When you're working with (A,T), then a point is not a real number, but a sequence of real numbers (i.e. an element of A). So you take two points and see if you can separate them by open neighborhoods, but a point is a sequence. Thus you pick two sequences satisfying your given condition and try to find open neighborhoods of them in T that separates them (or alternatively prove that this is impossible).

    How would you suggest to pick a real number x from A anyway? There are no real numbers in A.
  4. Apr 4, 2010 #3
    This is what i went ahead and did. I proved, for my particular space, that it is Hausdorff.

    What i think i confused it with, is that, i was thinking of A as if it were defined to consist of the terms of such sequences, instead of the sequences themselves.

    Thanks for clarifying it further.
  5. Apr 4, 2010 #4
    If you already defined a metric, say d, and T is the topology generated by it, then it's immediately Hausdorff. All metric topologies are Hausdorff, you don't have to consider the particulars of the space involved.

    This is a consequence of the triangle inequality: if you have two distinct elements x,y of A, no matter what they are, then d(x,y) = r > 0. Consider the open balls of radius r'<r/2, centred at x and y, respectively, and apply the triangle inequality to prove that these balls are disjoint.
  6. Apr 4, 2010 #5
    This is precisely what i did.

    But now when you bring it up, i realize that i have, indeed, proved the general case (that any metric topology is Hausdorff) rather than for my particular topology... since i never really made explicitly specific reference to my particular sequences, rather than just saying let x,y be any two such points(sequences in my case).
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