- Thread starter
- #1

I was just wondering why the subsets must be nonempty. Is it just convention/convenient or is it because it would violate something else?

Thanks!

- Thread starter Ragnarok
- Start date

- Thread starter
- #1

I was just wondering why the subsets must be nonempty. Is it just convention/convenient or is it because it would violate something else?

Thanks!

- Mar 10, 2012

- 834

I think it's just a matter of definition.

I was just wondering why the subsets must be nonempty. Is it just convention/convenient or is it because it would violate something else?

Thanks!

Certainly, if you don't follow this, many results about partitions will need to be tweaked.

There is nothing mathematically wrong in including the empty set also when considering partitions but that is not how it has been done.

- Feb 15, 2012

- 1,967

It is sometimes useful to consider partitions of the empty set itself; intuitively one feels that this partition should be unique, which is only possible if there are NO possible empty sets in a partition (convince yourself that this indeed comprises the only partition possible in the definition you gave for an empty set X).

- Admin
- #5

- Mar 5, 2012

- 8,780

For that to work, we need to be able to define a limit where the partition becomes infinitely fine grained, meaning each subset in the partition should be of size greater than or equal to $\varepsilon >0$.

In other words: not empty.

- Feb 15, 2012

- 1,967

Wrong kind of partition: the kind the OP is talking about might be a partition into singleton subsets, which would be fine....but such a partition would be rather unsuitable for integration.

For that to work, we need to be able to define a limit where the partition becomes infinitely fine grained, meaning each subset in the partition should be of size greater than or equal to $\varepsilon >0$.

In other words: not empty.