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Given a set $A$ is there a known way to find a topological space $X$ such that $|A|=|X|<w(X)$?

Here $w(X)$ is the weight of the topological space.

This is clearly impossible for finite sets $A$. We know it is possible for $A=\mathbb Z$ because of the answers to this question.

Is there a general construction for any infinite set? Is it even known if such a space exists for every infinite $A$? Has a contruction been found for $\mathbb R$ for example?

Thank you kindly.

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Let $U$ be a (free) ultrafilter on an infinite set $X$ which contains only sets of cardinality $|X|$ (that is, it is a normal ultrafilter). Put $\tau = U \cup \{\varnothing\}$; this is clearly a topology.

There is no base $B$ for $\tau$ with $|B|=|X|$ as every such base must have cardinality at least $|X|^+$ by a diagonal argument.

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  • $\begingroup$ It's not clear to me how to use a "diagonal argument" to prove that every uniform ultrafilter on $X$ has weight bigger than $|X|$. (That being said, I can prove that it's true for some uniform ultrafilters using independent sets, and this is good enough for an affirmative answer to the original question.) Am I missing something? $\endgroup$
    – Will Brian
    Commented Feb 14, 2017 at 20:39
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    $\begingroup$ @WillBrian To show that no family of $|X|$ sets in $U$ can be a base, enumerate the family as $(X_\xi)_{\xi<|X|}$ and then, by induction on $\xi$, pick $x_\xi$ and $y_\xi$ in $X_\xi$, distinct from each other and from earlier choices. Then each $X_\xi$ meets both the set of $x_\xi$'s and its complement, so the $X_\xi$'s can't be a base for $U$. Then it follows that they can't be a base for the topology either. $\endgroup$
    – Andreas Blass
    Commented Feb 14, 2017 at 22:01
  • $\begingroup$ @AndreasBlass: Thanks for clearing that up. It is indeed a (pretty straightforward) diagonal argument, and now I'm not really sure what I was thinking when I wrote my comment. $\endgroup$
    – Will Brian
    Commented Feb 15, 2017 at 14:44
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    $\begingroup$ Shouldn't "normal" be "uniform" here? I have always seen "normal" mean "closed under diagonal interesections" $\endgroup$
    – Danielle Ulrich
    Commented Feb 16, 2017 at 0:40
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Look at this article by I. Juhasz and K. Kunen for a construction of a Hausdorff space $X$ with $d(X)=\kappa$, $|X|=2^{2^\kappa}$ and $w(X)=2^{2^{2^\kappa}}$, where $\kappa$ is any cardinal given in advance.

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