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Second-countable space, first-countable spaceĬontractible space, locally contractible space Sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact Kolmogorov space, Hausdorff space, regular space, normal space Metric space, metric topology, metrisable space
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convenient category of topological spaces.
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Open subset, closed subset, neighbourhoodīase for the topology, neighbourhood base See also differential topology, algebraic topology, functional analysis and topological homotopy theory So ( x, y, z ) : | x + y + z | = 1 ( x, y, z ) : | x + y + z | = 1 is not a subspace.Topology ( point-set topology, point-free topology) Well, this is actually a non-linear function, and we can show that it is not a subspace pretty easily through a counterexample, by showing that the set is not closed under addition. What about things that are not subspaces? For instance: ( x, y, z ) : | x + y + z | = 1 ( x, y, z ) : | x + y + z | = 1 Therefore, as the subset defined by ( x, y, z ) : 2 x + 3 y = z ( x, y, z ) : 2 x + 3 y = zis a subspace of the volume defined by the three-dimensional real numbers. So if we multiply through α α: 2 α x + 3 α y = α z 2 α x + 3 α y = α zĪnd this is always true, since it is just a scalar. Again, we need to prove this by analysis: Now for being closed under multiplication.Which is always true, so we are closed under addition Now, if we recall that 2 x 1 + 3 y 1 − z 1 = 0 2 x 1 + 3 y 1 − z 1 = 0 and 2 x 2 + 3 y 2 − z 2 = 0 2 x 2 + 3 y 2 − z 2 = 0, so what we really have is: 0 = − 0 0 = − 0 Is it closed under addition? To work this out, we can take any two v 1 ∈ S v 1 ∈ S and v 2 ∈ S v 2 ∈ S and show that their sum, v 1 + v 2 ∈ S v 1 + v 2 ∈ S:.Is the zero vector a member of the space? Well: 0 = 2 × 0 + 3 × 0 0 = 2 × 0 + 3 × 0, so yes, it is.So for instance, valid subspace for a three dimensional space might be z = 2 x + 3 y z = 2 x + 3 y, which is a plane:Īnd does ( x, y, z ) : z = 2 x + 3 y ( x, y, z ) : z = 2 x + 3 y satisfy our three propositions? Well, we can prove this with a bit of analysis:
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