# What would this open set in R look like?

1. Jun 10, 2010

### lunde

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.

2. Jun 10, 2010

### Hurkyl

Staff Emeritus
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.

3. Jun 11, 2010

### Landau

As a well ordening of the reals requires the axiom of choice, I don't think one could imagine how the set looks like.

4. Jun 11, 2010

### HallsofIvy

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.

5. Jun 11, 2010

### Hurkyl

Staff Emeritus
That one's guaranteed -- every ordinal has a successor. It's the reverse that can fail: limit ordinals do not have a predecessor.