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Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Asaf: indeed, I added it for clarification, thank you.
Jul
14
revised Can the indexing set of a cartesian product have the cardinality of the continuum?
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Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Manos: I'll add on the non-emptyness of the product to the answer, as per Asaf's suggestion.
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Manos: No, I really meant exists. Not everything you define naïvely is a set ("the set of all sets" is not a set! Russell's paradox), you actually have to prove from the very axioms of set theory that you have defined something which really is a set... For a classical example, if $A$ and $B$ are sets, then as I said above, $A\times B:= \{(a,b):a\in A,b\in B\}$ is a set, but you have to prove it. If you describe a set by $\{x:\varphi(x)\}$ where $\varphi$ is a formula, you have to prove it exists: it may not exist (again, Russell's paradox: $\{x: x\notin x\}$ is not a set).
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Manos: you're welcome. I added some formality to the answer.
Jul
14
revised Can the indexing set of a cartesian product have the cardinality of the continuum?
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Jul
14
answered Can the indexing set of a cartesian product have the cardinality of the continuum?
Jul
14
comment How to add compound fractions?
I'd like to point out, that, however, $x= \frac{1}{\frac{1}{x}}$ only when $x\not=0$.
Jul
14
comment Mixing Problem: gallons in tank at a time t
@Zev: good edit, I had read "galois in tank at a time $t$" which was slightly disturbing! :D
Jul
14
awarded  Civic Duty
Jul
13
revised What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
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Jul
13
comment What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
@Asaf: have you read my answer thoroughly? I'm not saying that $A^I$ makes no sense for infinite $I$ -- on the contrary! If you are referring to the last line, I'm saying that defining $A^I$ as a set of tuples, for infinite $I$, makes no sense a priori. But of course, as a set of functions, sure.
Jul
13
comment What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
@Asaf: they can't be defined on a first year undergraduate course :)
Jul
13
comment What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
@confused: I've added a couple of things which may be nice to you. To think of $\mathbb{N}^2$ as pairs of natural numbers is the same that thinking $\mathbb{N}^2$ as the set of functions $2\to \mathbb{N}$, since $2=\{0,1\}$ in ordinary set theory. This is easily generalized.
Jul
13
revised What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to?
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Jul
13
comment What does “extend linearly” mean in linear algebra?
@Jack: of course not! My bad.
Jul
13
revised What does “extend linearly” mean in linear algebra?
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Jul
13
revised What does “extend linearly” mean in linear algebra?
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Jul
13
comment What does “extend linearly” mean in linear algebra?
@Jack: sorry! It is a common shortcut for "linearly independent". I edited it out.
Jul
13
revised What does “extend linearly” mean in linear algebra?
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