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I will illustrate my question in the case of the definition of vector spaces.

It is custom to define a vector space in the following way: "Let $K$ be a field. Then a $K$-vector space is a set $V$ together with the operations $+:V\times V\rightarrow V$, $\cdot:K\times V\rightarrow V$, such that the following holds. $[...]$". In "$[...]$" the axioms of the vector space are specified. Since the statement like "a set $A$ together with the operations $+,\cdot$" usually means we consider the triple $\left(A,+,\cdot\right)$, since this "glues" $A$ with the operations $+,\cdot$" together".

Thus this definition might be translated into a slighty more formal statement to say that "a vector space is a tuple $\left(V,+,\cdot\right)$ such that the following holds: $[...]$". What I'm not ok with, is that in the tuple $\left(V,+,\cdot\right)$ the field over which we take our vector space to be, isn't present. Therefore we actually don't have all the information we need, to state the axioms, if we don't implicitly assume that we already know about which field we talk.

We may know about which underlying set of the field we talk about, since in the mapping $\cdot:K\times V\rightarrow V$, the underlying set $K$ of the field $K$ (notice that the symbol $K$ is overloaded) is present, but we don't know about $+_{K},\cdot_{K}$, since these definitely aren't present in $\left(V,+,\cdot\right)$.

To give an example, from $\left(\mathbb{R}^{n},+,\cdot\right)$ we can't deduce if it is a vector space over $\mathbb{R}$ or $\mathbb{C}$ or some other field.

1) Is this slightly more formal translation of the more informal definition correct ?

To repair this situation, one solution I can think of would be to say "a vector space is a tuple $\left(V,+,\cdot,\left(K,+_{K},\cdot_{K}\right)\right)$, where $\left(K,+_{K},\cdot_{K}\right)$ is a field and the following holds: $[...]$".

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I think the last sentence of your question is your answer. –  Gerry Myerson Oct 7 '12 at 12:01
    
GerryMyerson Ok, but are there other (more elegant) solutions maybe ? Would my solution withstand general scrutiny of the math.SE ? –  temo Oct 7 '12 at 13:52
    
I have not been authorized to speak on behalf of math.SE. I think Peter Smith has given a good answer. –  Gerry Myerson Oct 7 '12 at 22:31

1 Answer 1

up vote 4 down vote accepted

To take a trite example, consider the Weiner-Kuratowski implementation of tuples. The story goes: an ordered pair $(a, b)$ is the set $\{\{a\}, \{a, b\}\}$, and then an $n+ 1$-tuple is an ordered pair of its head and its $n$-tuple tail.

All very familiar. But despite the familiarity, it would be very odd to say that this implementation gets things "right", is uniquely "correct". That won't wash even at initial level of the ordered pairs: after all, $\{\{a\}, \{a, b\}\}$ is not ordered, in the general case doesn't have two members, and other equally good implementations of pairs are possible.

Still, sure, for many purposes the Kuratowski implementation of tuples works just fine (though not for all -- if I recall, NF-istes need a different implementation). And the point generalizes. In giving set-theoretic implementations of mathematical notions, we aren't seeking unique "correctness" (whatever that might possibly mean). We just look for something or other that will work well enough, given our current local purposes, whatever they are.

So, if it is already settled what the relevant field $K$ supplying the scalars is, then there is nothing wrong with saying that the vector space is the triple $(V, +, \cdot)$ i.e. on the usual K-W implementation the pair $(V, (+, \cdot))$. But the pair $((V, +), \cdot)$ would do just as well. If you want to supply the field $K$ explicitly, then your suggested quadruple would do fine. But you wouldn't be going wrong if you wrote $(V, \times, \cdot, K, +_K, \cdot_K)$ or clumped things some other way. You are, to repeat, just looking for something that works conveniently, given your purposes: there is no issue of unique correctness here.

I might add, though, that is pretty unclear to me what playing with tuples buys us that we don't more clearly get from the initial informal talk of a set equipped with operations, etc. What purposes does the "formalisation" serve? What real benefits does it give us?

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Btw, who are NF-ists ? –  temo Oct 9 '12 at 17:57
    
@temo Enthusiasts for NF -- Quine's 'New Foundations' system of set theory. –  Peter Smith Oct 9 '12 at 18:30
    
But the benefit of $\{\{a\}, \{a, b\}\}$ over $(a, b)$ is that it doesn't depend on our geometric interpretation of the symbols $(a, b)$ with $a$ being on the left and $b$ being on the right. In that sense, the Kuratowski pair exists in a more abstract way than the ordinary human picture that is $(a, b)$. –  AmadeusDrZaius Mar 26 at 18:18
    
@temo: NF-istes are set theorists stout enough of heart to keep producing research that no one reads. No matter what, all six of us march on :D –  Malice Vidrine Jun 11 at 22:27

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