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Alberto Takase's user avatar
Alberto Takase's user avatar
Alberto Takase's user avatar
Alberto Takase
  • Member for 10 years, 1 month
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6 votes
1 answer
1k views

How to show $\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}\cong \mathbb{Z}/n\mathbb{Z}$

6 votes
1 answer
202 views

Notions of Infinity in ZF

4 votes
1 answer
296 views

Without the Axiom of Choice, $\aleph_0<2^{\aleph_0}$ implies $\aleph_1\le 2^{\aleph_0}$?

4 votes
3 answers
1k views

Transitive Closure of a Well-Founded Relation is Well-Founded (without Axiom of Choice)

4 votes
1 answer
176 views

Do we need the axiom of replacement (ZFC) to define a product of structures (Universal Algebra)?

4 votes
0 answers
151 views

supremum of additive functions is additive

3 votes
1 answer
166 views

Burali-Forti paradox for cardinals

3 votes
2 answers
449 views

Halmos' Naive Set Theory Cardinal Arithmetic Exercise

2 votes
2 answers
76 views

Dedekind-Infinite Set s.t. $|A\times A|=|A|$

2 votes
1 answer
571 views

What is a construction (in mathematics)?

2 votes
1 answer
135 views

Proving integrability given local Lp integrability

2 votes
1 answer
99 views

Boolean Algebra map extension theorem

1 vote
0 answers
88 views

K-Theory of $C(X)$ for $X$ not totally disconnected

1 vote
1 answer
66 views

Describing the sup-closure within a poset without the Axiom of Choice

1 vote
1 answer
76 views

Terminology for dual notion of continuity

1 vote
3 answers
245 views

For all finite sets $A$ and $B$, $\{ f:B\to A \}$ is finite and its (finite) cardinality is $|A|^{|B|}$.

1 vote
1 answer
81 views

The function space from $n$ to $m$ and the exponent $m^n$ are equinumerous (proof)

1 vote
1 answer
130 views

Definition of Ordinal (w/ Axiom of Regularity) (problem 37, page 208, Enderton's Elements of Set Theory)

1 vote
1 answer
890 views

Enderton's Elements of Set Theory Scott's Trick Exercise (page 207 problem 31)

1 vote
1 answer
339 views

For each ordinal $\alpha$, $\alpha\le \aleph_{\alpha}$

1 vote
1 answer
115 views

Enderton's isomorphism type arithmetic

0 votes
2 answers
240 views

$R$ well-founded relation and $\forall y$, $\{x:xRy\}$ is finite implies $\forall y$, $\{x:xR^t y\}$ is finite (where $R^t$ is the transitive closure)

0 votes
1 answer
191 views

Given a collection of functions $f_i$ with the same domain, how to replace with values (w/o axiom replacement)

0 votes
1 answer
211 views

Equivalent Definition of Well-Ordered Set

0 votes
1 answer
129 views

Union is finite implies the collection is finite

0 votes
1 answer
270 views

Egoroff's theorem for non $\sigma$-finite measure spaces, but each $f_n$ is integrable