Def.: For a topological space $S$ with weight $w(S)=\kappa$, I define the grasp $g(S)$ to be the least infinite cardinal $\gamma$ such that $S$ has a base $\mathscr{B}$ with $|\mathscr{B}|=\kappa$ such that every open set is the union of at most $\gamma$ members of $\mathscr{B}$.

When I spoke to Franklin Tall (Prof. Emeritus, U.of Toronto) last fall, he said he hadn't heard of it. (His specialty is point-set topology and set-theory).

I can show the following by elementary means:

  • [R1] $g(S) \le w(S)$ (obviously).

  • [R2] If $E$ is a subspace of $D$ and $w(E)=w(D)$, then $g(E)\le g(D)$.

  • [R3] Let $D[\mu]$ be the discrete space of cardinality $\mu$, where $\mu$ is an infinite cardinal. If $w(S)=\mu$, then $g(S)\le g(D[\mu])$.

  • [R4] If $\mathscr{T}$ is the topology on $S$ then $|\mathscr{T}|\le w(S)^{g(S)}$ (obviously).

  • [R5] Let the infinite cardinal $\kappa$ have either the discrete topology or the order topology. Then $\operatorname{cf}(\kappa)\le g(\kappa)\le \kappa$, and if $\kappa$ is a strong limit cardinal then $g(\kappa)=\operatorname{cf}(\kappa)$. NOTE: This last part of [R5] is useful to distinguish the grasp from other topological cardinal functions.


  • [Q1] For a topology $\mathscr{T}$ on $S$, the grasp $g(S)$ is not necessarily the least infinite $\lambda$ such that $|\mathscr{T}|\le w(S)^ \lambda$, for if $\mathscr{T}$ is the discrete topology on $\omega _1$ and if $2^\omega = 2^{\omega _1}$, then $|\mathscr{T}|=\omega _1^ \omega$, but the grasp is $\omega_1$, not $\omega$. Is there an example like this in $\mathsf{ZFC}$?

  • [Q2] What are the possible consistent values for the grasp of the discrete space $D[2^ \omega]$ aside from the value $2^ \omega$?


  • [E1] The Sorgenfrey line has weight $2^ \omega$ and grasp $\omega$. So in [R3] we may have, by [R5], that $g(S) < \operatorname{cf}(\chi)\le \chi=g(D[w(S)])$.

  • [E2] By [R1] and [R2],the Nimitzky plane $P$ has $g(P)=g(D[2^ \omega])$.

Any further thoughts about this function will be appreciated.

  • $\begingroup$ If $S$ has a base $B$ such that $\lvert B\rvert\le k,$ then by definition of weight, we have $\lvert B\rvert=k,$ do we not? $\endgroup$ Commented Aug 17, 2015 at 18:55
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    $\begingroup$ With 815 reputation points you are no longer a beginner. At this level, you are expected to use LaTeX formatting. $\endgroup$
    – Alex M.
    Commented Aug 17, 2015 at 18:59
  • $\begingroup$ (1)It has been said that "There are no finite cardinals in point-set topology." The weight is the least infinite cardinal k such that there is a base of cardinal k or less. (2) Thanks for improving the layout.I still haven't learned very much LaTeX. I haven't been here long. $\endgroup$ Commented Aug 17, 2015 at 19:31
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    $\begingroup$ @user254665: This site has a MathJax tutorial. \\ The one obvious comment is that $h(X)\le g(X)$, where $h(X)$ is the height (= hereditary Lindelöf degree). $\endgroup$ Commented Aug 17, 2015 at 19:53
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    $\begingroup$ @user254665: Oops. The inequality in my earlier comment is backwards. $\endgroup$ Commented Aug 17, 2015 at 20:22


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