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Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
I suggest you take a look at en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory
Jul
14
comment Can the indexing set of a cartesian product have the cardinality of the continuum?
@Manos: great question. How do you know it? Axioms! For example, $\mathbb{N}$ is a set (axiom of infinity). Also, the power set of any set is a set (axiom of power set). A subset of a set is a set (axiom of comprehension). The range of a function is a set (axiom of replacement). (I'm being deliberately non-formal). You can prove from the usual axioms of ZF that $\emptyset$ is a set. So, we know there are some things which are sets by axioms, and then we know how to build more sets from them, by axioms which allow us some constructions.
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?
@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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