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Hausdorff spaces

  1. Jul 8, 2013 #1
    Hello everybody!

    It is known that in Hausdorff spaces that every sequence converges to at most one point. I'm curious whether this characterizes Hausdorff spaces. If in a space, every sequence converges to at most one point, can one deduce Hausdorff?

    Thanks in advance!
     
  2. jcsd
  3. Jul 8, 2013 #2

    WannabeNewton

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    Not necessarily. Let ##X## be an uncountable set with the cocountable topology ##\mathcal{T} = \{U\subseteq X:X\setminus U \text{ is countable}\}##. Assume there exist two distinct points ##p,p'## and two neighborhoods ##U,U'## of the two points respectively such that ##U\cap U' = \varnothing ##. Then ##X\setminus (U\cap U') = (X\setminus U)\cup (X\setminus U') = X##. But ##(X\setminus U)\cup (X\setminus U')## is a finite union of countable sets which is countable whereas ##X## is uncountable thus we have a contradiction. Hence ##X## is not Hausdorff under the cocountable topology.

    Now let ##(x_i)## be a sequence in ##X## that converges to ##x \in X## and let ##S = \{x_i:x_i\neq x\}##. This set is countable so ##U\setminus S## must be a neighborhood of ##x## in ##X##. Thus there exists some ##n\in \mathbb{N}## such that ##i\geq n\Rightarrow x_i\in U## but the only distinct element of the sequence that is in ##U## is ##x## so ##x_i = x## for all ##i\geq n## i.e. any convergent sequence in ##X## must be eventually constant under the cocountable topology. Hence limits of convergent sequences must be unique (the map prescribing the sequence must be well-defined).
     
  4. Jul 8, 2013 #3
    Great! Thanks a lot! The example you gave is very interesting since it has the same convergent sequences as the discrete topology.
     
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