I'm not getting this...like I have a pattern for it but UGH! I have to provide a proof for a general case that this satisfies the definition not setting k and m at a certain number. I'll screenshot the question and type out what I have so far. So, it's gonna take 5-10 min to get it set up
http://prntscr.com/knnk9s #11 is the problem. I already did #10 and it did satisfy the definition which is closed under complements, unions, and intersections.
http://prntscr.com/knnkjm
But now I have this long string and I have to use m =2 and k = 1 which that was the two sets A and B and only considering one point in either A or B which is probably what the k = 1 meant
so the symmetric difference is $A + B = ( A \backslash B) \cup (B \backslash A)$
and then by complement definition $A + B = ( A \cap B^{c}) \cup (B \cap A^{c})$
which would mean all points in set A or all points in set B
If I take the complement of that then it's $A + B = ( A \cap B^{c})^{c} \cup (B \cap A^{c})^{c}$
$A + B = ( A^{c} \cup B ) \cup (B^{c} \cup A)$
taking the complement back again will give me the original A+B and I think the m = 2 is the number of sets in total and k =1 the number of sets considered when taking the points?
So now like if m = 3 and k = 2 then I will have three sets A,B,C
which will become ABC with three cases if I were to take two of any of these sets A and B but Not C A and C but not B B and C but not A
$(A \cap B \cap C^{c}) \cup (A \cap B^{c} \cap C) \cup (A^{c} \cap B \cap C)$
so that would mean out of three sets only points from each of the two sets will be considered. it goes on and on like when m =4 and k =3. That's ABCD but then what I'm stuck is how to write an overall general proof of this since this has to work on every case and not when picking a specific number like what was done here
that's what I'm trying to get....like a clearer picture of this. When I did the previous three exercises 8-10 they were all related to the definition. Ohhh is it like induction
base case, k case, k+1 case?
like if k = 1 then we have to show that it belongs to at least one of them and it's it's 1 and not in 1+1=2...
but having that is kinda in F since it needs to be nonempty. And if A is in F then so is the complement A or $ \Omega \backslash A \in F$ and then it's closed under finite unions... hmmm
@DrewBrady I think what they are asking is a proof by contradiction .. suppose $\lim_{t \to \infty}$ don't agree for the limit of integral and integral of the limit .. then there is a sequence $t_n$ etc
atleast that was the essence of the C7 appendix they quoted
wait a sec... I think I might have some idea. Let $\displaystyle \cup_{i=1}^{n} \left(A_i \cap ( \cup_{j\neq i} A_j)^{c}\right)$ is in F The event space F contains both the empty set and $\Omega$ . Since F is nonempty, then $A_{i} \in F$ and the complement is also in F. So by setting $A_{i}=A$ and $A_{j} =A^{c}$ for $i \geq 2$ then F also contains the union $\Omega = A \cup A^{c}$
Oh I see, what you're saying. It's a "hint" because this textbook does not actually define lim_{x \to a} f(x) as defined iff lim_{n to \infty} f(x_n) for all (x_n) distinct -> a.
they use an eps-delta definition which is equivalent to my characterization (which is how I learned it in analysis)
ok.
just a question of semantics really then it looks like.
@usukidoll it reads .. point is in $A_i$ bot not in any other $A_j$ (which is why I took the union of the rest and differenced it out of $A_i$) .. then you take union on all $i$
it's 10:30 am here // bbl .. @usukidoll now it's easy to generalize to $k$-case ..
ooh x-x . So I have to use this definition http://prntscr.com/knnuv1 for this. in t his case $\displaystyle \cup_{i=1}^{n} \left(A_i \cap ( \cup_{j\neq i} A_j)^{c}\right)$
@DrewBrady I don't get what 1-11 is talking about but it's related to 1-10 which I already did because I used the definition to prove that it's in F which is a collection of events prntscr.com/knnwiu
it said something about previous case is the symmetric difference with m = 2 and k =1
In a $2 × 4$ rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid?
My attempt :
I found a formula somewhere,
Number of rectangles are $= m(m+1)n(n+1)/4 = 2\times4\times3\times5/4 = 30$.
Can you please explain in formal way?
Update :...
whoa whoa whoa whoa .. so it will go like Suppose we have a multi-index such that $1,...,m$ is $(a_{1},...,a_{k})$ where $a_{i}$ are distinct and $a_{i} \in \{1,...,m \}$
ah ok so so far it seems like we have Set I = $\{1,...,m \}$ of size k and then set J = $\{ a_{1}, a_{2},...,a_{k} \}$ and then the $a_{i} $ are distinct elements of set I which was 1,2,...,m
It's going to be in the sigma algebra because I is finite, and B_J is in the sigma-algebra since it is finite set operations to sets in the sigma algebra.
Based on our conversation, it seems unlikely that you followed the entire discussion, so my strong recommendation to you would be (1) make the question a post on math.stackexcahgne, and post the link here, so I can explain the discussion we just had in greater detail.
I still got confused on some parts. It was this one huge index of two things and then I sort of see that since A_{j} was in F then so was A_{k}$ because that's the complement or something like that. x.x
yeah I think I should post the question on main but what do I call it?
In a $2 × 4$ rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid?
My attempt :
I found a formula somewhere,
Number of rectangles are $= m(m+1)n(n+1)/4 = 2\times4\times3\times5/4 = 30$.
Can you please explain in formal way?
Update :...
Typically factors taken into account when choosing which of the two questions should be duplicate target are: age, quality of question, quality of answers.
I would consider good title which is easy to find as a part of "quality of the question" criterion. And high number of views might indicate that this one is easy to find.
So many cases of spelling their way through a word only to come up with something completely different than the word really is because of 3 exceptions occurring in that word
Because I said so :) (you should prove it yourself. It is a good exercise to show that this holds for all prime powers of odd primes)
For powers of $2$, we get an extra factor of order $2$ and the rest is cyclic.
To be fair, I don't actually recall the full argument, so I would need to work a bit on it. The idea is to use a generator for the prime power one below to construct one above.
@AnotherJohnDoe Was that one comment really that bad? We try not to be too serious here, and the question you asked is very hard to answer without knowing much more about you as a mathematician
Munkres is a hard but useful book. It is best to spend more than half a year on it to ensure you fully grasp the principles as topology can be quite non intuitive for infinite dimensional spaces
i will say, unless you have good coursework related reasons, do it slowly, better to understand than to be confused
@TobiasKildetoft well I knew at least functional spaces will need more attention on when first came across those since there are no diagrams to guide you and the proofs will be mostly algebraic
Hi, I'm trying to write a proof for $(A\cup B)\cup C=A\cup(B\cup C)$. I expressed both sides with respect to definition of $\cup$ of sets. And now, it seems obvious to me because $(p\lor q)\lor r$ has the same truth value of $p\lor(q\lor r)$. But I feel like I couldn't show it clearly. Any ideas?
In a $2 × 4$ rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid?
My attempt :
I found a formula somewhere,
Number of rectangles are $= m(m+1)n(n+1)/4 = 2\times4\times3\times5/4 = 30$.
Can you please explain in formal way?
Update :...
