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Given the expression

$s\in\{(u_0,u_1)\in \Bbb R \times\Bbb R^3\},$

what is s?

  • a single vector in $\Bbb R^4$ consisting a concatenation of the elements of $u_0\in \Bbb R$ and $u_1\in \Bbb R^3$
  • a set of two vectors in $\{u_0, u_1\in\Bbb R^4\}$
  • a subset consisting of a element $u_0\in\Bbb R$ and an element $u_1\in\Bbb R^3$
  • a matrix in $\Bbb R^{u_0\times3}$

Also, how would the following be expressed: "s is either a real number or a 3-vector given that the real number equals the norm of the 3-vector"

Is this along the right lines? $s\in\{u_0\in\Bbb R, u_1\in\Bbb R^3\,|\,u_0=||u_1||_2\}$

mel
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    Your first bullet point is correct. One can always view $\mathbb R^{n}$ as being $\mathbb R^m \times \mathbb R^k$ where $m+k = n$. More generally, ${ (x,y) \in A \times B}$ means pairs $(x,y)$ with $x \in A, y\in B$. – ah11950 Apr 03 '14 at 11:18
  • I'm assuming by your most recent edit, you're asking how one might express the set of all pairs of vectors in $\mathbb R^4$. By the reasoning above, this is of course $\mathbb R^4 \times \mathbb R^4 = {(u,v) \mid u \in \mathbb R^4, v \in \mathbb R^4}$ – ah11950 Apr 03 '14 at 11:30
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    Too many edits for me to keep up with... $s \in {u_0, u_1 \in \mathbb R \times \mathbb R^3}$ doesn't make sense. Looks like you need to go over set builder notation. When we talk about the Cartesian product $A \times B$ of two sets, then an arbitrary $x \in A \times B$ is of the form $x=(a,b), a \in A, b\in B$. So ${u_0, u_1 \in \mathbb R \times \mathbb R^3} = {u_0, u_1 \in \mathbb R^4} = \mathbb R^4$ (since "$\mathbb R^3 \times \mathbb R = \mathbb R^4$" – ah11950 Apr 03 '14 at 11:36
  • @ah11950 Thank, that helped. Now that I know it is called "set builder notation" I am sure I can find the rest of the way on my own. Many thanks. – mel Apr 03 '14 at 11:39
  • third point is exact. There is trivial bijection with first point – Sylvain Biehler Apr 03 '14 at 11:41
  • @Bilou06 So the (...) do not necessarily denote pairing but can also denote a subset? – mel Apr 03 '14 at 11:43
  • Saying subset is a little dangerous; elements of Cartesian products $A\times B$ are ordered pairs, whereas subsets (presumably of $A\cup B$ in this context) are unordered. This only really matters if you have $s\in A\times A$; the elements $(a_1,a_2)$ and $(a_2,a_1)$ of $A\times A$ are different, but the $2$-element subsets ${a_1,a_2}$ and ${a_2,a_1}$ of $A$ are the same. – mdp Apr 03 '14 at 12:33

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