# The class indexed by real numbers is a set?

1. Mar 29, 2012

### julypraise

Let $\mathcal{S} = \{S_{i}:i \in \mathbb{R} \}$ where $S_{i}$ is a set. Then $\mathcal{S}$ is a set? Or, can this notation make sense in some way?

2. Mar 30, 2012

### DonAntonio

I can't see why you think S couldn't be a set, as long as each $S_i$ is...What did you have in mind?

DonAntonio

3. Mar 30, 2012

### Whovian

A set can be of anything. Even {Lincoln, Charizard, {Fish Fingers, Custard}} is a set. Your set is, therefore, valid. Note that its cardinality is one of the alephs, I don't know which one.

4. Mar 30, 2012

### Hurkyl

Staff Emeritus
Because the $S_i$ are sets, this is valid set builder notation defining a class $\mathcal{S}$. And by the axiom of replacement and the fact $\mathbb{R}$ is a set, the class $\mathcal{S}$ is indeed a set.

Not anything. There isn't, for example, a set of all sets that don't contain themselves!

5. Mar 30, 2012

### Whovian

True. Sorry for poor wording. "Almost anything" would've been a better wording.

6. Mar 30, 2012

### SteveL27

That's a perfectly valid set. But note that by convention, indexing by $i$ typically indicates indexing over the natural numbers. For clarity, it would be better to write

$\mathcal{S} = \{S_{\alpha}:\alpha \in \mathbb{R} \}$

which provides readers with an indication that we are indexing over a set other than the natural numbers.

7. Mar 30, 2012

### julypraise

You know, the concept of indexing in my mind (in my intuition) is kind of a countable process. But then now the index set is a continuum. So I thought it might not be possible; I mean this kind of indexing might not be possible by ZFC.

8. Mar 30, 2012

### DonAntonio

I see. But, as already noted by others, it is possible to index by means of any set, no matter its cardinality.

In ZFC we can even use Zermelo's Well Ordering Theorem to well order ℝ and then well-order the so indexed sets, if so wanted...

DonAntonio

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