countable basis造句
例句與造句
- So, these are T2 spaces with countable basis, yet non metrizable.
- The vector space spanned by the countable basis is the set of all "'finite "'linear combinations of the basis vectors.
- If there's no countable basis for the vector space of all sequences of reals, then there's no vector-space isomorphism.
- I know that a space X is Lindel鰂 if every open cover has a countable subcover and that X is a second countable space of its topology has a countable basis.
- Right now I don't have that answer either & mdash; it seems as if there should be some obvious way to show there's no countable basis.
- It's difficult to find countable basis in a sentence. 用countable basis造句挺難的
- A "'Schauder basis "'or "'countable basis "'is similar to the usual ( Hamel ) basis of a vector space; the difference is that for Hamel bases we use linear combinations that are " finite " sums, while for Schauder bases they may be " infinite " sums.
- Now if you recall how the proof of " regular + countable basis-> normal " goes, we can apply the following theorem : if A and B can be covered by two countable collections { Un } and { Vn } respectively, and that the closure of each Un is disjoint from B, and the closure of each Vn is disjoint from A, then A and B can be covered by disjoint open sets.
- I can see how the polynomials, which must have a maximum degree for any polynomial, would be isomorphic to the space of sequences with only finitely many nonzero terms, with the obvious bijection between coefficients and terms in the sequence, and this must be contained in \ mathbb { R } ^ { \ mathbb { N } }, but I have a feeling the latter may have an uncountable basis whilst the former has the obvious countable basis e i = ( 0, 0, . . ., 1, 0, 0, . . . ) with the 1 in the i-th position-but how would I show such a thing if so?