# Vectors/Cartesian products

1. Jul 6, 2012

### Cinitiator

1. The problem statement, all variables and given/known data
Let's say we have a notation:

Does it mean that x is an L-tuple? Also, if it does, does it have to have L members?
If it x isn't an L-tuple, does it mean that x belongs to an L-tuple? That is, x is inside of an L-tuple: (1, x, ... L)? And again, does it mean that that L-tuple has to have exactly L members?

2. Relevant equations

3. The attempt at a solution
Googling

Last edited: Jul 6, 2012
2. Jul 6, 2012

### micromass

Staff Emeritus
Yes.

Yes.

For example, typical members of $\mathbb{R}^6$ are x=(1,2,3,6,4,2). An element like (1,2,3) is not an element of $\mathbb{R}^6$.

3. Jul 6, 2012

### tiny-tim

Hi Cinitiator!
Yes.
Yes!!

A Cartesian product space, of L individual spaces, only has elements that are L-tuples.

If the individual spaces are all different, that's fairly obvious … it's the same even if the individual spaces are identical.

4. Jul 6, 2012

### janny

Hi Cinitiator,

Does it mean that x is an L-tuple?

No, x $\in$ ℝL+ usually means that x is a column vector with L rows with real entries where the subscript + requires that all its entries are positive.

http://en.wikipedia.org/wiki/Column_vector

A tuple is also a kind of array but such that the listed objects are ordered horizontally. It is usually notated like this:
L-Tupel: (n1,n2,...,nL)

http://en.wikipedia.org/wiki/Tuple

Also, if it does, does it have to have L members?

Yes, exactly L entries (or alternatevely members)

Last edited: Jul 6, 2012