What would this open set in R look like?

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Well order the real numbers, let {a_n}_{n \in S_{\Omega} } be the the singleton sets of odd numbers in the well order (i.e. skip a number, grab a number, skip a number, grab a number). Since the real numbers are T_1 then singleton sets are closed. \mathbb{R} as a topological space is closed by definition. Let U := \mathbb{R} \bigcap_{n \in S_{\omega} } {a_n} . Then U is closed and hence \mathbb{R} \backslash U is open.

Is my logic correct here? It just seemed strange to me that this set would be open. Does anyone know what this set would look like? Thanks.
 
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Your construction is somewhat confusing, and in my best guess at your intent, U is the empty set.


Incidentally, what are you doing about infinite ordinals? I suppose you could decree limit ordinals to be even, and extend the notion of being even/odd from there.
 
As a well ordening of the reals requires the axiom of choice, I don't think one could imagine how the set looks like.
 
More importantly, while any set can be "well ordered", that does NOT imply that, given one number, there exist a "next" number. That is only possible for "countable" sets and the real numbers are not countable.

You could ask this question of the rational numbers but then I think that Landau's comment is valid.
 
HallsofIvy said:
More importantly, while any set can be "well ordered", that does NOT imply that, given one number, there exist a "next" number.
That one's guaranteed -- every ordinal has a successor. It's the reverse that can fail: limit ordinals do not have a predecessor.
 
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