Is it true that $(x,y,z)=((x,y),z)=(x,(y,z))$?

Question

Is there any difference between $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$, $\mathbb{R} \times \mathbb{R}^2$ and $\mathbb{R}^2 \times \mathbb{R}$?

Two of them have two coordinates, a scalar and a vector. The first one is clearly the 3-dimensional space $\mathbb{R}^3$. But what about the other two?

Is there any topological difference?

Answer 1

As mentioned in the comments we usually view these as formally different objects: An element of $\Bbb R^3$ is a triple of elements in $\Bbb R$, whereas an element of $\Bbb R \times \Bbb R^2$ is a pair whose first component is a point of $\Bbb R$ and whose second is an element of $\Bbb R^2$. There are however, canonical bijections between these sets, e.g., $$\Bbb R^3 \leftrightarrow \Bbb R \times \Bbb R^2, \qquad (x, y, z) \leftrightarrow (x, (y, z)) ,$$ and these bijections will respect most familiar structures you might put on these sets. For example, if we give $\Bbb R^k$, $k = 1, 2, 3$, the usual vector addition and scalar multiplication rules, these bijections are vector space isomorphisms.

A choice of topology $\tau$ on $\Bbb R$ determines product topologies $\tau^k$ on $\Bbb R^k$, $k > 1$, and for any choice of $\tau$, e.g., the above bijection is a homeomorphism $(\Bbb R^3, \tau^3) \cong (\Bbb R \times \Bbb R^2, \tau \times \tau^2)$, where $\tau \times \tau^2$ is the product topology of $\tau$ and $\tau^2$. For general choices of topologies on $\Bbb R, \Bbb R^2, \Bbb R^3$, however, they will not be homeomorphisms.

Answer 2

The difference on the level of sets between these three is the one you point out: the elements differ by parenthesis. So, all three are different as sets. As topological spaces they are homeomorphic. As James suggests, take $\phi: (\mathbb{R}^2)\times\mathbb{R}\to \mathbb{R}^3$ to be the function $((x,y),z)\mapsto (x,y,z)$. We can take it as granted that this is a bijection. Now all you need to check is that $\phi$ and $\phi^{-1}$ are continuous. To show $\phi$ is continuous, you need to show that for every open set $U$ of $\mathbb{R}^3$, $\phi^{-1}(U)$ is open in the product topology $(\mathbb{R}^2)\times\mathbb{R}$. One way to do this is to show that an arbitrary open set $U\subset\mathbb{R}^3$ is actually a union of smaller open sets that are $z-$thickened slices parallel to the $xy-$plane.