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Suppose we define a topology on the real numbers as that obtained by defining open sets to be those such that the complement is finite. How does one show that this topology cannot be obtained from a distance function?

Thanks in advance.

p.s: This is not a homework question. I am trying to self study point set topology.

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    The term is _metrizable_. In a standard reference like Munkres you will find many topological properties that metric spaces satisfy, and a space that does not satisfy one of these properties is not metrizable.2010-08-22
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    Hausdorff is one in a sequence of _separation axioms_ a topological space can satisfy (which are discussed in detail in Munkres). Metric spaces satisfy a rather strong separation property: they are normal (http://en.wikipedia.org/wiki/Normal_space) as well as Hausdorff. So any topological space which does not satisfy these kinds of separation properties is not metrizable.2010-08-22
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    You might like the book *Counterexamples in Topology*.2010-08-22
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    As a footnote to Qiaochu's comment, what you describe is precisely the Zariski topology on the real line,* which is highly pathological from the perspective of analysis or algebraic topology (when most of the spaces one cares about are much more than Hausdorff). *Though normally working with algebraically closed fields is nicer.2010-08-23

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Your topology is called the finite complement topology of the real line: http://en.wikipedia.org/wiki/Finite_complement_topology . To show that it's not Hausdorff, try to intersect any two open sets -you don't even need the points of the Hausdorff definition-, remembering the rule

$$ (X \backslash A ) \cap (X \backslash B) = X \backslash (A \cup B) \ . $$

Having any Topology standard book at hand, like Munkres, might help your self study. For instance, I'm quite shure that the proof that every metric space is Hausdorff can be found there.

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Try showing that metric topologies are Hausdorff and that this topology isn't.

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    What does Hausdorff mean?2010-08-22
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    A topological space $X$ is called Hausdorff if for any $p,q\in X$ there are open sets $U_p$ and $U_q$ such that $p\in U_p$, $q\in U_q$ and $U_p\cap U_q = \emptyset$. In words, a topological space is called Hausdorff if any two distinct points can be separated by disjoint open sets.2010-08-22
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    @Shibi, if you google for "Hausdorff" the very first result is the Wikipedia page you'd have to read to find out what Hausdorff means :)2010-08-23
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As mentioned above, the topologies you want are those which are not metrizalbe. There are several metrication theorems that are given in terms of separation properties of the spaces.

  1. A space is first countable if every point has a neighborhood basis
  2. A space is second countable if it has a countable basis

All metric spaces are first countable, so if your space is not first countable then it is not metrizable.

Probably the most commonly known theorem is Urysohn's metrization theorem: every second countable, regular, Hausdorff space is metrizable. Hausdorff just means that points can be separated by open sets and regular means that (nonempty) closed sets and points can be separated by open sets.

A more powerful condition is that of normality. A space is normal if disjoint closed sets can be separated by open sets. It turns out that all metrizable space are normal.

At first glance, these non-metrizable spaces just seem like pathological examples, but some are very important. For example, the Zariski topology is not metrizable, but is essential to the field of algebraic geometry.