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I'm having a little trouble grasping the definition of a topological space in terms of the open-sets. Suppose that we're defining a topology on the real line, in theory could we arbitrarily select the elements of the open-sets, so long as the set of all open sets satisfies the properties of a topological space?

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    So, you have two open sets which consist each of two points, right? Then you can generate a [discrete topology](http://en.wikipedia.org/wiki/Discrete_space) with this. That is, if you add a number of extra axioms like translating an open set is still giving an open set, but that is quite natural to do on the reals.2010-12-20
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    Sorry, I just deleted that to make the question more clear. What I'm current confused about is whether an arbitrary set can be defined as "open" without satisfying other properties such as convergence?2010-12-20
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    Yes, you can define any set to be open, provided when you take the collection of sets that you defined to be open and you make unions and finite intersections, these sets also be open. But of course, not every choice will lead to interesting topologies. The discrete and the trivial topology are not really interesting from the viewpoint of analysis.2010-12-20
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    I'm not sure what you're asking, but if you're asking whether a set of subsets of the real line defines a topology if and only if it satisfies the conditions in the definition of a topology, then the answer is yes.2010-12-20
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    @niale Could you please clarify what do you mean by "convergence"?2010-12-20
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    I think what he wants is to see how topology provides the basis for notions such as limits in analysis.2010-12-20
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    Yes arbitrary set can be defined as open, but the catch is, that you need to define all open sets. The convergence is then determined by the open sets you have. If you have non-interesting open sets (trivial topology for example), your convergence can be non-interesting. For trivial topology for example any element in topological space is a limit for any sequence.2010-12-20
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    Thanks very much all. Things are making sense now. @adrian I was referring to convergence in a metric space.2010-12-20

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