Download Definitions of set operations and relations

Comments on the proofs (symbols and logics)
Dear all,
I think most of you did well on HW1. But you could probably improve the
style and logic of the proof even if you got a perfect score. The followings are a
couple of formal things you all probably aware of, or at least conceptually.
Definitions of set operations and relations
The following definitions of set operation seem perfectly clear at the first glance,
but you sometimes have trouble finding out how to prove certain questions
simply because you don’t quite recall their definition. So here it is. If A, B are
both sets, define A ∪ B, A ∩ B, A \ B as
A ∪ B = {x : x ∈ A or x ∈ B}
A ∩ B = {x : x ∈ A and x ∈ B}
A \ B = {x : x ∈ A and x ∈
/ B} = A ∩ B c
Also, here’s a definition of set relation ⊆ .
A ⊆ B means x ∈ A =⇒ x ∈ B.
and A = B is defined as A ⊆ B and B ⊆ A. So if you want to prove a statement
A ⊆ B, start by picking any element x ∈ A, then try to prove that x ∈ B. If
you succeed, then the proof will be done. If you want to prove A = B, do both
A ⊆ B and B ⊆ A though sometimes A ⊆ B are granted. For that case, you’ll
prove reverse inclusion. Solution of first problem in HW2 is this case.
Similarly A * B means x ∈ A ; x ∈ B or equivalantly, there is some x ∈ A
which is not an element of B. If you want to prove A ⊆ B, try to pick x that is
exclusively in A, i.e. x ∈ A but ∈
/ B. If A * B were our assumption, your first
line of proof should be ‘Let a ∈ A \ B’. Problem 9 of first HW relates this.
Here is an example relating application of this set-theoretic definition. Suppose V be a vector space such that V = span{v1 , v2 } and U1 = {av1 : a ∈ F },
U2 = {bv2 : b ∈ F } be subspaces of V where {v1 , v2 } is linearly independent.
Then V = U1 ⊕ U2 .’ To prove this argument, you need to show V = U1 + U2
and U1 ∪ U2 = 0. For the first part, what does it mean by two sets are equal?
it means V contains U1 + U2 and V is contained in U1 + U2 . First part is
automatic, since V is the whole set. For the other direction, we need to show
V ⊆ U1 + U2 . That is by definition above, we need to prove ‘for any v ∈ V ,
then v is an element of U1 + U2 ’. Now, we use the condition V = span {v1 , v2 }
to show any v ∈ V is written as v = a1 v1 + a2 v2 for some a1 , a2 ∈ F . Then
a1 v1 inU1 , a2 v2 ∈ U2 to conclude v ∈ U1 + U2 since v is sum of vector in U1 and
that in U2 . For converse, we want to show U1 ∩ U2 = 0. So as before, we need
both inclusion. 0 ∈ U1 and 0 ∈ U2 so 0 ∈ U1 ∩ U2 is easy. Conversely, we want
U1 ∩ U2 ⊆ 0 that is, by definition, pick any vector v ∈ U1 ∩ U2 . Then we want
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v = 0. To see this, since v ∈ U1 ∩ U2 , v ∈ U1 and U2 so v = a1 v1 and v = a2 v2
for some a1 a2 ∈ F . Then we have a1 v1 − a2 v2 = 0 and since {v1 , v2 } is linearly
independent, a1 = a2 = 0. Therefore v = 0v1 = 0 so we’re done.
Symbols
I agree that writing a full sentence in a proof every time is quite tedious, so
we tend to abbreviate certain things with symbols. But it makes the reader
(or grader in fact) confused if you write a wrong symbols. So here is couple of
things I noticed.
a ∈ A implies ‘a is an element of A’, while B ⊆ A implies B is a subset of
A. They have totally different meaning. If R is the set of real numbers, 0 ∈ R
and {0} ⊆ R make sense, while 0 ⊆ R is totally nonsense, because 0 is not a
set. {0} ∈ R symbollically makes sense, but is just not true. If we denote P (R)
as collection of all subsets of R, then {0} ∈ P (R) makes sense and is true.
Second important thing you should know is that when you want to say first
line implies the second, try to use double arrow =⇒. In mathematics, p −→ q
stands for a statement ‘If p, then q’ while p =⇒ q stands for the statement
p −→ q is true. Similarly a symbol p ⇐⇒ q stands for ‘p holds if and only if q
does’ or ‘p is equivalant to q’. So for example, in problem 1,
√
−1 + 3i
ω=
2√
=⇒ 2ω + 1 = 3i
=⇒
···
=⇒ ω 3 = 1
It is okay (though not quite encouraged) to write in opposite direction like
ω3 = 1
⇐=
···
√
−1 + 3i
⇐= ω =
2
But if you write up in a opposite direction, you must indicate the first line is the
one we want to prove while the last line is our assumption. One way of indicate
this is a symbol ⇐=
Overall proofwriting
On the first line of proof, please define a relavant symbols which may appear
in the proof. If it is mentioned in the problem, you don’t need to do so. For
example, if problem goes like ‘If U1 and U2 are subspaces, then U1 ∪ U2 is not a
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subspace unless U1 ⊆ U2 or U2 ⊆ U1 .’ Then you can freely speak of Ui because
the problem mentioned about Ui . But if the same problem goes like ‘The union
of subspace is not a subspace unless one contains the other.’ Then your first line
of proof should be ‘Let U1 , U2 be subspaces such that U1 * U2 and U2 * U1 .’
Sometimes reader could recognize it by context but you sometimes miss some
conditions that would be crucial in your proofs, which may results in no points
at all.
Also, please make full use of variables. It will make your proof much clean.
For example, in the first problem of HW2, we substitute vi − vi+1 for wi and
vn for wn . It’s not a big deal. Read the proof again without substituting wi
and see how much you need to write more. Also in HW2 problem 8, we used a
term like ‘standard ordered basis’, which makes you don’t need to write up all
these five vectors as a coordinate. (you could actually omit the i.e. part). It
may be used as to distinguish the same thing defined in a different ways. For
example, consider you want to prove the following argument : span {vi } is the
minimal subspace containing all vi . Then you may speak it as let v be in the
minimal subspace containing all vi . Then the minimal subspace containing all
vi must contains all linear combinations of vi . (blah blah) so minimal subspace
containig all vi equals span {vi }. It is quite stressful to write the same thing
‘minimal subspace containing all vi ’ over and over again. Rather, we could write
‘Let V1 = span{vi } and V2 be a minimal subspace containing all vi . Then since
all linear combination of vi is in V2 , (blah blah) so V2 = V1 .’ This sure saves
time, so be aware of it.
Next thing I should mention is that please distinguish between what you
want to prove and what you already got. One way of distinguish is to use a lot
of transitions. I’m not the one who’s good at English, but you could write better
proof with a few combination of transition things. When you want to remind
what you need to prove, use like ‘I claim that ...’. If you have some sufficient
condition to prove, use ‘It suffices to show that ...’ or ‘We only need to show
that ...’ in front of the thing you want to prove. Without these transition,
just to put a simple formula means that the formula is true, which sure confuse
the reader. Also, please distingush between you pick something and you set
something. For example, in problem 13 and 15, we want a counterexample, so
we ‘set’ Ui to violates the statement. If we were to prove it, we ‘pick’ any Ui
and make the argument works.
Finally, I see some of you wrote down something that was not asked or
unnecessary. This may happen because you couldn’t fully understand the questions or what to prove. One way of avoding it is first to briefly sketch what
you gonna do, rather to write down a proof. For example, in problem 13, it
says prove or disprove the following. So if you want to prove it, then you sketch
your proof like ‘Let U1 , U2 , W be subspaces such that U1 + W = U2 + W . blah
blah. Therefore U1 = U2 ’. If you want to disprove it, sketch your proof like,
‘Take U1 = ...U2 = ...W = .... Then somehow U1 + W = U2 + W but U1 6= U2 .’
After that, fill out the ‘blah blah’ or ‘...’ and put some reasoning instead of
‘somehow’.
I planned to upload these comments at least earlier this week, so I apologize
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for the lateness. The followings are some side remarks which has less importance.
This comment become much longer than I first expected. I was afraid it would
distract from the key things, so I moved less important comments to the side
remarks. If you have time, please continue reading.
Side remarks
In some linear algebra book, you can sometimes see a symbol like 0 ⊆ R, as if
0 were a set. (and I also did it in this article. Where did I?) In that context,
0 stands for a zero vector space, not a zero element. Zero vector space is, as a
set, {0}, so 0 = {0} ⊆ R now makes sense. Though two symbols 0 looks the
same, they have different meaning, one being a set and the other is an element
in R. In mathematics, if a single symbol has more than one meaning, it is
called ‘abuse of notation’. This is perfectly fine if the context shows things
clearly. You’ve already seen that 0 stands both for scalar 0 ∈ F and zero
vector(additive identity) 0 ∈ V . 0 ∈ F n is (0, 0, · · · , 0), so it’s not quite equal
to 0 ∈ F .
What’s the difference of term set ‘operation’ and ‘relation’ ? by term operation, we mean the result is again a set while by term relation, the result is true
or false statement.
The proof of exercise in set theoretic part is to practice how those formal
definition applied to certain problem, so the proof is unusually kind. You should
know this flow of ideas but in a real proof, you don’t need to write all like this.
These set-theoretic implications are hidden in the proofs. Maybe it’s fun to find
out which statements and flow of ideas were implicited in the solutions.
Sometimes we deals with really many collection of sets. Let’s say [a, b] stands
for closed interval {x : a ≤ x ≤ b}. Now we want to consider ∩n∈N [0, 1/n]
where collections are taken for various n ∈ N . So what does that mean? You
can probably think, ‘okay, it means x ∈ [0, 1] and x ∈ [0, 1/2] and x ∈ [0, 1/3]
and ... and ... and ...’. You’re right, but due to some reason in logics, you’re
not allowed to take a infinite argument. So you may think it as
∩[0, 1/n] = {x : x ∈ [0, 1/n] for some n}
and the definition above meets our concept. You’re allowed to write it less
formally as [0, 1] ∩ [0, 1/2] ∩ · · · . But how about ∩α∈R>0 [0, α]? This family of
intervals are no longer countable, so we cannot list it like I1 ∩ I2 ∩ · · · . Also, you
cannot list your argument like ‘x ∈ [0, 0.3624] and x ∈ [0, π] and x ∈ [0, π + e]
and so on’. The only possible way is to define it in the above sense. So here is
the definition: If {Wα } are any collection of sets, ∪Wα and ∩ Wα is defined as
∪Wα = {x : x ∈ Wα for some α}
∩Wα = {x : x ∈ Wα for all α}
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