Perfection is a local property of a topological space : a space is perfect if and only if every point in the space admits a basis of neighborhoods each of which is perfect in the subspace topology.
32.
Equivalently, a space " X " is locally regular if and only if the collection of all open sets that are regular under the subspace topology forms a base for the topology on " X ".
33.
The same argument as before shows that the subspace topology on Z is not equal to the induced order topology on Z, but one can show that the subspace topology on Z cannot be equal to any order topology on Z.
34.
The same argument as before shows that the subspace topology on Z is not equal to the induced order topology on Z, but one can show that the subspace topology on Z cannot be equal to any order topology on Z.
35.
There is, on the other hand, a theorem that says that ( in the subspace topology ) there is an analytical bijection between small enough neighborhoods of the identity in the group and the origin in the Lie algebra respectively.
36.
If " X " is an affine algebraic set ( irreducible or not ) then the Zariski topology on it is defined simply to be the subspace topology induced by its inclusion into some \ mathbb { A } ^ n.
37.
This corresponds to the regular topological definition of continuity, applied to the subspace topology on A \ cup \ lbrace p \ rbrace, and the restriction of " f " to A \ cup \ lbrace p \ rbrace.
38.
Note however, that if a space were called locally normal if and only if each point of the space belonged to a subset of the space that was normal under the subspace topology, then every topological space would be locally normal.
39.
Which can also be identified with the set of basis for the open sets of the product topology are cylinder sets; the homeomorphism maps these to the subspace topology that the Cantor set inherits from the natural topology on the real number line.
40.
Under the subspace topology, the singleton set { 1 } is open in " Y ", but under the induced order topology, any open set containing 1 must contain all but finitely many members of the space.
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