Why do we look at non-flat geometries in Cosmology?

In Cosmology we use the Robertson-Walker-Metric which follows from the cosmological principle & mathematics. This metric leaves three cases for a possible curvature (or geometry) of space (not spacetime). The curvature is describes in this metric with a curvature parameter $k$ which can be -1,0,1. Often different models are solved for special cases (like $k=-1$ and we just have matter etc.).

Since measurement reveal that $k=0$ (flat universe) why do we deal in cosmology with those other (physically irrelevant) cases like k=-1 and k=+1 if we know that k=0?

• Measurements can only tell us that the universe is close to flat today (and in the past). They can't tell us that it is exactly flat and they can't predict what the future will be. – CuriousOne Jan 28 '16 at 11:25
• @CuriousOne So you mean that the curvature paramter changes with time? – Thomas Elliot Jan 28 '16 at 12:03
• I don't know if it does. I especially don't know what it will do in the future. How would I or anybody? – CuriousOne Jan 28 '16 at 16:04

Very often we don't. We just stick to $k=0$, because for most practical purposes, this is fine. But all measurements contain uncertainties. The latest Planck results (Parade et al. 2015) combined with observations of the baryonic acoustic oscillations yield$^\dagger$ $\Omega_k = 0.000\pm0.005$. We don't know whether the curvature on much larger scales than the observable Universe is flat. If $|\Omega_k| \lesssim 10^{-4}$, we won't know (Vardanyan et al. 2009).
$^\dagger$$\Omega_k$ is the fraction of the Universe's total energy content "contained" in curvature, defined through it's present-day's value $\Omega_{k,0} = kc^2/H_0^2$, and $\Omega_k(a) = \Omega_k/a^2$, where $a$ is the scale factor describing the relative size of the Universe, defined to be $1$ today.
• @ThomasElliot and igael: I wouldn't really call it bounds. The curvature parameter $k$ is simply defined so that it can only assume one of the three values $-1$, $0$, or $+1$, for a scale factor $a$ defined to be $1$ today. On the other hand, $\Omega_k$ can assume any value. If $\Omega_k=0.111111243545$, then $k=+1$, and if $\Omega_k=-3.2\times10^{-16}$, then $k=-1$. I edited the answer a bit to describe this relation better. – pela Jan 28 '16 at 14:04