# Tag Info

## Hot answers tagged set-theory

28

Set theory and category theory are both foundational theories of mathematics (they explain basics), but they attack different aspects of foundations. Set theory is largely concerned with "how do we build mathematical objects (or what could we build)" while category theory is largely concerned with "what structure to mathematical objects have (or could ...

17

As others have mentioned, the collection of all metric spaces and the collection of all topological spaces both form proper classes, so we can't reason about their size using cardinality. A similar question that can be answered is: given a set $X$, are there more topologies on $X$ than metrics on $X$? To this end, given a set $X$, let $T(X)$ be the set of ...

6

$\mathscr{T}$ and $\mathscr{M}$ are proper classes in ZFC, not sets. (They're also proper classes in set theories that can talk about proper classes explicitly — von Neumann-Godel-Bernays set theory and Morse-Kelley set theory.) In fact the two classes can be put in 1-1 correspondence: see the answer by Q the Platypus, which points out that the ...

6

The class of all sets is bijective to the class of all metric spaces. Because every set can be injectively mapped to the a metric space via the trivial metric and since every metric space is a pair of sets there is a injective map from the the class of all metric spaces back to the class of all sets. Likewise since the class of all topological spaces is a ...

5

"Width" here is meant in the informal sense of how big each "level" $V_\alpha$ of the cumulative hierarchy is. The point is that, if $W[G]$ is a forcing extension of $W$, then they have the same number of levels, but each level of the extension is at least as large as each level of the original: $$(V_\alpha)^W\subseteq (V_\alpha)^{W[G]}.$$ Moreover, as soon ...

4

Here's an answer for the case of finite spaces: there are countably many finite topological spaces up to homeomorphism (because each finite set admits only a finite number of topologies). There are uncountably many finite metric spaces up to isometry (because just for a space with two points $x$ and $y$ you have uncountably many choices for $d(x, y)$). This ...

4

Yes. Suppose not, look at $X^c$ which is of positive measure. By Lebesgue Density Theorem, there exists $\sigma, \tau\in 2^{\omega}$ (WLOG might assume they have the same length) such that $X$ and $X^c$ has measure $>\frac{1}{2}$ above $\tau, \sigma$ respectively. By hypothesis, $X$ has measure $>\frac{1}{2}$ above $\sigma$ too. But then above ...

3

You're confusing well-ordered with well-orderable. There is no uncountable well-ordered chain in $\mathcal P(\Bbb N)$, because if there was such $\{A_\alpha\mid\alpha<\omega_1\}$, map $\alpha$ to the least $n$ in $A_{\alpha+1}\setminus A_\alpha$, and you got yourself an injection from $\omega_1$ into $\omega$. Something which you cannot get with or ...

2

Without additional hypotheses, not too much actually. It cannot be a Borel set, since uncountable Borel sets have a copy of the Cantor set and thus they have cardinality continuum. It cannot be of positive measure, since if $A$ is of positive measure, $A-A=\{a-b\mid a,b\in A\}$ contains an interval, and thus $A$ had to be of size continuum. Similarly if it ...

2

The claim is not true in ZF alone. It is consistent with ZF that there are infinite (i.e. not finite) sets which are Dedekind finite (i.e. not equinumerous with any proper subset). For such sets, adding even one element changes the cardinality. Two further comments: The amount of choice needed here is very little; countable choice, for instance, is more ...

2

The other answers here deal (correctly) with cardinality questions about the classes you're asking about. I don't think that's the way to ask the question that seems to interest you. Perhaps you want to know in what sense there are more topological spaces than metric spaces. If that's your question then ... Every metric space is a topological space in a ...

1

The measure theoretic part was answered, so let me complement it by answering the choice related question. The axiom of countable choice is needed on a far more fundamental level when you talk about measure theory. It is consistent that the real numbers are a countable union of countable sets. In that case there is no $\sigma$-additive Borel measure ...

1

All the definitions given seem to agree that $0$ is always in the diagonal intersection. The sets $\{\xi\in\kappa: \xi\in\bigcap_{\alpha<\xi}X_\alpha\}$ and $\bigcap_{\alpha\in\kappa}(X_\alpha\cup\{\xi: \xi\le\alpha\})$ are the same set; in particular, $0$ is in the latter, since for every $\alpha\in\kappa$ we have $\alpha\ge0$, so \$0\in\{\xi: ...

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