Setminus
Axis
1) Is there a positive equivalent, that is a separate symbol from the plus
sign, to the set minus sign "\" ? \setplus is rejected by Latex.
Currently I've got, for my set L(n),
L(f(n))= L(n) + m in [1,t].
I want m to be a single integer that is to be incorporated into L.
2) I wonder if the following is acceptable, given that, by assumption,
|L(n)|+1 = |L(f(n))| :
L(f(n)) \ L(n) = r
?
Here I want r to a single value, a member of L(n); but I wonder if it would
be an abuse of convention to write it as anything other than {r}?
With thanks in advance.
Brian M. Scott
<zed...@hotmail.com> wrote in
<news:OVGTk.45078$%D2.1...@newsfe18.ams2> in
alt.algebra.help:
> Two quick questions, if I may:
> 1) Is there a positive equivalent, that is a separate
> symbol from the plus sign, to the set minus sign "\" ?
> \setplus is rejected by Latex.
You're probably talking about the union symbol, LaTeX \cup.
> Currently I've got, for my set L(n),
> L(f(n))= L(n) + m in [1,t].
> I want m to be a single integer that is to be incorporated into L.
You probably want
L(f(n)) = L(n) \cup {m},
but that can't be immediately followed by 'in [1, t]'. What
are you trying to say with that last bit?
> 2) I wonder if the following is acceptable, given that, by
> assumption, |L(n)|+1 = |L(f(n))| :
> L(f(n)) \ L(n) = r
> ?
> Here I want r to a single value, a member of L(n); but I
> wonder if it would be an abuse of convention to write it
> as anything other than {r}?
Yes, it would. You really do need to write L(f(n)) \ L(n) =
{r}.
Brian
Axis
"Brian M. Scott" <b.s...@csuohio.edu> wrote in message
news:75ys2dnkdmrt.1n07zw4d0wnrz$.dlg@40tude.net...
> On Sat, 15 Nov 2008 21:11:12 -0000, Axis
> <zed...@hotmail.com> wrote in
> <news:OVGTk.45078$%D2.1...@newsfe18.ams2> in
> alt.algebra.help:
>
>> Two quick questions, if I may:
>
>> 1) Is there a positive equivalent, that is a separate
>> symbol from the plus sign, to the set minus sign "\" ?
>> \setplus is rejected by Latex.
>
> You're probably talking about the union symbol, LaTeX \cup.
>
>> Currently I've got, for my set L(n),
>
>> L(f(n))= L(n) + m in [1,t].
>
>> I want m to be a single integer that is to be incorporated into L.
>
> You probably want
>
> L(f(n)) = L(n) \cup {m},
>
> but that can't be immediately followed by 'in [1, t]'. What
> are you trying to say with that last bit?
>
I've got a set [1,t] and I want to arbitrarily pick a member from it to add
to the set L(n) such as to make L(f(n)). Actually, I can see that it would
be best to write |L(f(n)|=|L(n)|+1.....
Brian M. Scott
<zed...@hotmail.com> wrote in
<news:tzWTk.90639$mr4....@newsfe19.ams2> in
alt.algebra.help:
>> You probably want
Then say so!
Let m in [1, t] be arbitrary, and let L(f(n)) =
L(n) \cup {m}.
> Actually, I can see that it would be best to write
> |L(f(n)|=|L(n)|+1.....
Not if you're trying to specify the construction of L(f(n)).
The statement |L(f(n))| = |L(n)| + 1 merely says that
L(f(n)) contains one more member than L(n) and implies no
relationship between the members of these two sets.
Brian