# Why is physical space equivalent to $\mathbb{R}^3$?

Why is physical space equivalent to $\mathbb{R}^3$, as opposed to e.g. $\mathbb{Q}^3$?

I am trying to understand what would be the logical reasons behind our assumption that our physical space is equivalent to $\mathbb{R}^3$ or 'physical straight line' is equivalent to $\mathbb{R}$ .

The set of reals $\mathbb{R}$ is a basically an algebraically constructed set, which is nothing but the completion of $\mathbb{Q}$, the set of rationals. For reference see here http://en.wikipedia.org/wiki/Construction_of_the_real_numbers. Now my question is what is the reason behind our approximation of the physical space by this abstract set. Why is this approximation assumed to be most suitable or good approximation?

• I suppose it is because we assume that space is continuous. If it is continuous each dimension can be assigned a one to one correspondence with a real number in the set. Now if we ever find that space is quantized, as some are trying to prove, then the continuity assumption breaks down and another set will have to be found to represent space in a one to one correspondence. The correspondence with a mathematical set is necessary if we ever want to calculate anything. – anna v Sep 24 '12 at 12:11
• Is this a question about dimensionality or about continuous vs discrete or both? – user11547 Sep 24 '12 at 12:12
• @Hal Swyers: about continous vs discrete ? why only these two ? Cant there be some other choice ? – pritam Sep 24 '12 at 12:15
• That depends on the context. When you are dealing with complex numbers, it is certainly feasible to have discrete magnitudes and continuous phases, or discrete phases and continuous magnitudes. The reason I was asking is that there is a history of debate over whether spacetime itself is discrete or continuous, which is a distinctly different debate than the number of spacetime dimensions. – user11547 Sep 24 '12 at 12:22
• Related: physics.stackexchange.com/q/20822/2451 and links therein. – Qmechanic Sep 24 '12 at 12:50