Final answer:
To prove m is not a 2-dimensional smooth manifold at q = (0.5, 1, 1), we must show that there is no local homeomorphism from a neighborhood of q to an open set in ℝ^2. This involves demonstrating that no such mapping can be both continuous and continuously invertible, or that m does not have the necessary smoothness at q.
Step-by-step explanation:
To prove by definition that a set m is not a 2-dimensional smooth manifold at the point q = (0.5, 1, 1), we need to refer to the definition of a smooth manifold. A smooth manifold of dimension k at a point p is a topological space that is locally homeomorphic to ℝk, meaning around every point p, there exists a neighborhood that is smoothly isomorphic (via a homeomorphism) to an open set in ℝk.
To disprove that m is a smooth manifold at q, we must show there is no such local homeomorphism that maps a neighborhood of q to an open set in ℝ2. This would usually involve demonstrating that no such mapping exists that is both continuous and continuously invertible, or that the set does not possess the necessary smoothness properties at the point in question. Without specific information about the set m, we can't give a precise explanation, but the proof would proceed by attempting to construct a mapping and showing it fails to satisfy the criteria for smoothness at q.