# Tag Info

11

If $C$ is a concrete category i.e. a category equipped with a faithful functor $U : C \to \mathsf{Set}$, then the free $C$-object over a set $X$ is by definition an object $F(X)$ which satisfies the universal property $\hom(F(X),A) \cong \hom(X,U(A))$, naturally in $A \in C$. If $C=\mathsf{Set}$ and we choose (what else?) $U=\mathrm{id}$, then clearly we ...

5

You can't construct such example. The fact that it is only consistent and not provable is exactly the issue here. Not only you can't construct one, you can't prove that one exists when the axiom of choice fails. These remarks, at the introductory level, are meant to guide the reader and suggest that a choice-free proof cannot be found. This is used to both ...

4

In abstract category theory morphisms are not required to be functions: we are given an arbitrary class, regarding its elements as arrows, equipped with domain and codomain information and an associative operation. If we want, we can specify so that for a funtion $f:A\to B$, let ${\rm dom\,}f:=B$ and ${\rm cod\,}f:=A$ and the operation is defined as ...

4

There are several ways to define regularity. You can do it using order preserving functions; or you can do it using partitions. I like working with the partitions, for this proof. $\kappa$ is regular if every partition of $\kappa$ into less than $\kappa$ parts, there is at least one part of size $\kappa$. For example $\aleph_\omega$ is not regular, ...

2

There is exactly one morphism from $X$ to $\emptyset$, namely the unique function from $\emptyset$ to $X$.

2

Because $\aleph_{\alpha+1}^{\aleph_\beta}=2^{\aleph_\beta}$. Because $\kappa$ is a singular cardinal exactly when there is an unbounded function from some $\lambda<\kappa$ to $\kappa$. Because each function is bounded, each function is in fact from $\omega_\beta$ into some $\gamma<\omega_{\alpha+1}$ (e.g. $\gamma$ taken as the successor of the ...

2

If by definable you mean "explicitly constructible by chaining a finite string of logical symbols" then clearly the size of the set of all definable numbers is countable. In particular, this means that most transcendental numbers can't be written down explicitly as formulas. However, if we allow for second order definability, then we can reach higher ...

2

No, $\beta_0^\alpha=\beta$, so $\beta_1^\alpha$ is the ordinal $\min(C_\beta\setminus\alpha)$: it’s the smallest member of $C_\beta$ that is greater than or equal to $\alpha$. It exists because $C_\beta$ is cofinal in $\beta$, and it’s necessarily less than $\beta_0^\alpha=\beta$, since $C_\beta\subseteq\beta$. If this is greater than $\alpha$, repeat the ...

1

My comment might be viewed as facetious, but it really wasn't intended that way. The point is that the nature of what's accepted as a definition of a "number" really has a lot to do with the answer to this question. If we accept this: "The set of all real numbers between $1$ and $2$." as an adequate characterisation of every real number in that interval, ...

1

Suppose $X$ is a metric space with a two valued diffused Borel measure. Let $\mathcal{U} = \bigcup \{\mathcal{U}_n : n < \omega\}$ be a basis for $X$ where each $\mathcal{U}_n$ is a disjoint family of open sets - See here. Let $\langle x_i : i < \kappa \rangle$ list $X$. For each $i$, let $B_i$ be a open ball around $x_i$ whose measure is zero. Let ...

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