0

A question stems when I reread definitions of basis of topology. I know that every topology is a basis of itself, but what theorems are there to check the cardinality of basis that a topology can have?

As for my attempt, it seems so Munkres doesn't discuss this in the second chapter and google/ SE search doesn't find me any easy results :/

tryst with freedom
  • 11,538
  • For most spaces, there are lots. Lots and lots. Presumably, you want a cardinality rather than "number?" For example, for any metric space such that for every $x\in X$ and $r>0$ there is a $y$ such that $d(x,y)=r,$ there is a distinct basis for any subset of the positive real numbers with $0$ in the closure. That is at least $P(P(\mathbb N))$ cardinality, and these are really simple bases. There are often many more. $X=\mathbb R^n$ with the usual metric has this property. – Thomas Andrews Aug 25 '22 at 18:41
  • 1
    Given any basis $\mathcal B$ of $X$ and an open cover $\mathcal U={U_{\alpha}}$ of $X,$ we can define a basis: $\mathcal B_{\mathcal U}={V\cap U\mid U\in\mathcal U, V\in\mathcal B}.$ So if we start with the basis of all open sets, then $\mathcal B_{\mathcal U}$ would be all open subsets of some $U_{\alpha}.$ If your open cover has no $U_i\subsetneq U_j,$ then the maximal elements of the basis are just the elements of the cover, so there is one basis per open cover when no cover element contains another. – Thomas Andrews Aug 25 '22 at 18:59
  • 1
    Over a finite topological space, each $x\in X$ has a minimal open subset $U_x$ containing $x,$ and any basis much contain each $U_x.$ In some cases, you might have $U_x=U_y.$ These are, essentially, indistinguishable points. If $\tau$ is your topology, then any subset of $\tau$ containing all the $U_x$ is a basis. If $k$ is the number of distinguishable points - that is, the number of points in $X/\sim$ where $x\sim y$ if $U_x=U_y,$ then $|\tau|\leq 2^k$ and number of bases is exactly $2^{|\tau|-k}.$ – Thomas Andrews Aug 25 '22 at 19:16
  • 1
    Given a topology $\tau$ on a set $X$ and basis $\mathcal B\subseteq\tau,$ any subset $\mathcal B\subset \mathcal B'\subseteq \tau$ is also a basis.

    The topological spaces we first encounter, the subspaces of $\mathbb R^n,$ all have a countable basis, and most have $\tau$ uncountable, so there are at least $2^{\mathfrak c}$ bases.

     – Thomas Andrews Aug 25 '22 at 19:33
  • @Thomas Andrews: There are at most $2^{\mathfrak c}$ many bases for the usual topology on $\mathbb R^n$ because each basis is a subset of the set of open sets, a set of cardinality $\mathfrak c.$ Near the beginning of my answer to Confusion Regarding Munkres's Definition of Basis for a Topology (see Question) I gave a description of $2^{\mathfrak c}$ many bases for the usual topology of $\mathbb R$ such that each basis contains only open intervals (this can also be easily modified for ${\mathbb R}^n$ with open balls or open cubes). – Dave L. Renfro Aug 25 '22 at 20:41

0 Answers0