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huangxienchen@xxxxxxxxx schrieb:
>
> Would you say that the solution set to
>
> 0 = 0 * a ( where a is a "given" constant)
>
> is topologically equivalent to
>
> 0 = 0 * x (where x is a variable)
>
> ???
No, why should I?
The first equation has no unknowns, hence there is nothing to solve at
all.
The second equation has one unknown.
In general, the solution set of an equation (or a set of equations)
with n unknowns is a set of n-tuples. It's a bit tricky to say what a
set of 0-tuples should be, but apparently there is only one 0-tuple:
(). And there are only two sets of 0-tuples: {} and {()}.
If you insist that the first equation should be viewed as an equation
in 0 unknowns,
then the set of solutions is (for any given [real?] a) the set {()}.
And, as has been stated often enough, when working over the reals, the
set of solutions to the second equation is R. R, for example with
standard topology, is not homeomorphic to {()} (which allows only one
topology).
> Certainly, whatever these solutions are, they are not equivalent to the
> null set of solutions which you'd get from 0 = 0.
In the same sense, viewing this as an equation in 0 unknowns, the set
of solutions would be {()}.
However, the set of solutions to 0 = 1 would be the null set {}.
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