[QUOTE=JohnnyMo1;40410571]
It doesn't imply that for general metric spaces. Spaces in what that does hold are sometimes called Heine-Borel spaces or said to have the Heine-Borel property.[/QUOTE]
While I hate to be that guy, Heine-Borel actually says that 'closed and bounded' => 'compact'
The converse is always true in metric spaces, which becomes fairly obvious if you use sequential compactness... (It is still possible to prove directly using the open cover definition, but you'd be basically proving that compact => sequentially compact again)
But yes, not every space has the Heine-Borel property, so certainly the two things aren't equivalent in general.
[QUOTE=Joey90;40411109]While I hate to be that guy, Heine-Borel actually says that 'closed and bounded' => 'compact'
The converse is always true in metric spaces, which becomes fairly obvious if you use sequential compactness... (It is still possible to prove directly using the open cover definition, but you'd be basically proving that compact => sequentially compact again)
But yes, not every space has the Heine-Borel property, so certainly the two things aren't equivalent in general.[/QUOTE]
Whoops. I read his post backwards. :v:
Hey guys, I'm getting my Math Foundations 30 credit to enter university, and am having some difficulties. Math is not my strong point, I'm definitely more of an English / History buff when it comes to getting high marks. My marks in math are pitiful.
I am currently doing Logic and Set Theory, if any of you are familiar with this, I am seeking a bit of help.
Basically, right now we are working on doing Venn Diagrams to determine certain things, an example right now that I have on my exam study sheet, is this:
148 students were surveyed, the info broke down in the following way:
87 students took physics
84 students took chemistry
68 students took biology
27 students took only physics and chemistry
15 took only physics and biology
33 took only chemistry and biology
18 did not take any science class.
I need to make a venn diagram to determine how many students took all three science classes.
I am struggling with this, though the whole thing was easy at first, these questions are baffling me.
Would it be possible for any of you to give me a step by step on how to approach and complete a question like this? I have an exam on this unit tomorrow and really would like to get a high mark in the class. My teacher is not the best at explaining these things... Any help is definitely very appreciated, thank you.
[editline]25th April 2013[/editline]
For anybody interested in helping me with the question I posted.. Here was what I did when tackling the problem myself.
I have a strong feeling that somewhere along the lines I made a mistake, but would like somebody who knows more about this math to give me a hand and point out where I went wrong...
I believe it may have been in subtracting the 18 students who did no science classes, but part of me says I did this right...
Here is my venn diagram and math on a poorly constucted ms paint product.
[IMG]http://i.imgur.com/xtLqT8t.jpg[/IMG]
[QUOTE=lifemonkey;40423556]Hey guys, I'm getting my Math Foundations 30 credit to enter university, and am having some difficulties. Math is not my strong point, I'm definitely more of an English / History buff when it comes to getting high marks. My marks in math are pitiful.
I am currently doing Logic and Set Theory, if any of you are familiar with this, I am seeking a bit of help.
Basically, right now we are working on doing Venn Diagrams to determine certain things, an example right now that I have on my exam study sheet, is this:
148 students were surveyed, the info broke down in the following way:
87 students took physics
84 students took chemistry
68 students took biology
27 students took only physics and chemistry
15 took only physics and biology
33 took only chemistry and biology
18 did not take any science class.
I need to make a venn diagram to determine how many students took all three science classes.
I am struggling with this, though the whole thing was easy at first, these questions are baffling me.
Would it be possible for any of you to give me a step by step on how to approach and complete a question like this? I have an exam on this unit tomorrow and really would like to get a high mark in the class. My teacher is not the best at explaining these things... Any help is definitely very appreciated, thank you.
[editline]25th April 2013[/editline]
For anybody interested in helping me with the question I posted.. Here was what I did when tackling the problem myself.
I have a strong feeling that somewhere along the lines I made a mistake, but would like somebody who knows more about this math to give me a hand and point out where I went wrong...
I believe it may have been in subtracting the 18 students who did no science classes, but part of me says I did this right...
Here is my venn diagram and math on a poorly constucted ms paint product.
[IMG]http://i.imgur.com/xtLqT8t.jpg[/IMG][/QUOTE]
Your diagram looks right, but the calculation at the side doesn't make much sense.
If you add up all the numbers on your diagram you have:
(45-x) + 27 + (24-x) + 15 + x + 33 + (20-x) + 18
= 182 - 2x
so you want to solve 182 - 2x = 148, which has solution x = 17.
[QUOTE=JohnnyMo1;40410571]I can think of some other examples if you'd like.[/QUOTE]
Thanks heaps, but I think I need more explanations about why things aren't compact from the topological definition. For example, X = (0,1) under the usual topology. I'm guessing a cover made from all balls of any radius such that the ball is totally contained in (0,1) couldn't have a finite subcover, but how can you show that? I know it has to do with the endpoints as [0,1] is compact, but to me I can't see why the ball around x=1/2 with radius 1/2 doesn't form a subcover for any cover of X.
[QUOTE=ThisIsTheOne;40425689]Thanks heaps, but I think I need more explanations about why things aren't compact from the topological definition. For example, X = (0,1) under the usual topology. I'm guessing a cover made from all balls of any radius such that the ball is totally contained in (0,1) couldn't have a finite subcover, but how can you show that? I know it has to do with the endpoints as [0,1] is compact, but to me I can't see why the ball around x=1/2 with radius 1/2 doesn't form a subcover for any cover of X.[/QUOTE]
Clearly the ball centered at 1/2 with radius 1/2 disproves your belief that the cover by all balls of any radius wholly contained in X doesn't have a finite subcover, but not every possible cover of X contains the ball of radius 1/2 centered at 1/2. Here's a simpler cover which doesn't have a finite subcover: {(1/n+1,1):n is a natural number}. Every point in (0,1) is contained in one of those sets, but if you take only a finite number of them, pick the one with the highest n, and the subcollection only contains (1/n+1,1).
[QUOTE=Joey90;40425459]Your diagram looks right, but the calculation at the side doesn't make much sense.
If you add up all the numbers on your diagram you have:
(45-x) + 27 + (24-x) + 15 + x + 33 + (20-x) + 18
= 182 - 2x
so you want to solve 182 - 2x = 148, which has solution x = 17.[/QUOTE]
He just forgot about the 18 students and some careless miscalculation
I wish I was good at this stuff, thank you for your help.
So after doing the math, this is what my equation looked like, my answer was 17, like you said it should be.
(45-x) + 27 + (24 - x) + 15 + 33 + (20-x) + x + 18
182 - x = 148
182 - 3x +x = 148
182 - 2x = 148 (subtracting 182 from both sides now)
2x = 34 (now dividing 2, right?)
x = 17
my math kind of confuses me, but i got the right answer.
[QUOTE=JohnnyMo1;40426110]Clearly the ball centered at 1/2 with radius 1/2 disproves your belief that the cover by all balls of any radius wholly contained in X doesn't have a finite subcover, but not every possible cover of X contains the ball of radius 1/2 centered at 1/2. Here's a simpler cover which doesn't have a finite subcover: {(1/n+1,1):n is a natural number}. Every point in (0,1) is contained in one of those sets, but if you take only a finite number of them, pick the one with the highest n, and the subcollection only contains (1/n+1,1).[/QUOTE]
That makes sense, thanks. Do you have any books you recommend for topology at an introductory level?
[QUOTE=ThisIsTheOne;40425689]Thanks heaps, but I think I need more explanations about why things aren't compact from the topological definition. For example, X = (0,1) under the usual topology. I'm guessing a cover made from all balls of any radius such that the ball is totally contained in (0,1) couldn't have a finite subcover, but how can you show that? I know it has to do with the endpoints as [0,1] is compact, but to me I can't see why the ball around x=1/2 with radius 1/2 doesn't form a subcover for any cover of X.[/QUOTE]
Yeah, a sort of intuitive (but obviously not rigorous) way of seeing why an open or unbounded set is not compact (in a metric space!):
If it's open, you must have some sort of open 'edge' somewhere, so you can take sets which get closer and closer to that 'edge', so no finite number will actually 'reach' the edge, but if you have the full infinity of them then they will. (This doesn't work for closed 'edges' since for those you actually need a particular set which does cover the edge). The protypical example is the one you gave, and the sets that Johnny suggested.
If it's unbounded, you can have sets which cover areas further and further away from the first, a finite number of them can only cover a finite distance away but since it's unbounded this won't get everything. (Again, if it is bounded this won't work since you actually need to reach the boundary with some set). The prototypical example here is R, where you can take sets (-n,n).
These are special cases of looking at when a sequence doesn't have a convergent subsequence - often it's either because the sequence converges to a point not in the space (the first example) or because it goes off to infinity (the second). And as such when working with metric spaces it is usually easier to work with sequential compactness.
This stuff doesn't really work in more general Topological spaces, since things can be much more weird, constructing the open covers/finite subcovers is different in each case.
[QUOTE=lifemonkey;40426264]I wish I was good at this stuff, thank you for your help.
So after doing the math, this is what my equation looked like, my answer was 17, like you said it should be.
(45-x) + 27 + (24 - x) + 15 + 33 + (20-x) + x + 18
182 - x = 148
182 - 3x +x = 148
182 - 2x = 148 (subtracting 182 from both sides now)
2x = 34 (now dividing 2, right?)
x = 17
my math kind of confuses me, but i got the right answer.[/QUOTE]
It's correct, but I'm not really sure what the second line is about... You should just go straight from the first to the third (or even straight to the fourth). The second line is just false :/
[QUOTE=lifemonkey;40426264]I wish I was good at this stuff, thank you for your help.
So after doing the math, this is what my equation looked like, my answer was 17, like you said it should be.
(45-x) + 27 + (24 - x) + 15 + 33 + (20-x) + x + 18
182 - x = 148
182 - 3x +x = 148
182 - 2x = 148 (subtracting 182 from both sides now)
2x = 34 (now dividing 2, right?)
x = 17
my math kind of confuses me, but i got the right answer.[/QUOTE]
Well, let's see if we can't make it a touch better for you. 148 students were surveyed; so all together, all of the students must sum to 148.
87 took physics, so all of the physics students must sum to 87.
You're then given that 27 students are taking physics and chem, so place 27 in the intersection of PHY and CHM. Now subtract your total physics students by 27 to get 60.
Then, you're told 15 took PHY and BIO, so put the 15 in the intersection of PHY and BIO, and subtract 60 by 15; 45.
Lastly, we do not yet know how many students had all three, so we place an x (any free variable) in the For all Intersection, and subtract x from 45; (45-x).
Therefor, going backwards to check, we have:
(45-x) + x = 45
45 + 15 = 60
60 + 27 = 87.
Wash, rinse, repeat for the CHM & BIO intersect, and you should be able to figure it out from there as you've done. Then simple arithmetic and algebra from there. Altogether, you should have all of that junk you have written which sum to 182 - 2x = 148, which yields x = 17.
Hope this helps a little, with a little bit more of a step-for-step approach. Let us know if you need some more help, or for the other guys, if I missed any steps that I just assumed was understood and skipped over.
[QUOTE=ThisIsTheOne;40428137]That makes sense, thanks. Do you have any books you recommend for topology at an introductory level?[/QUOTE]
I really like Munkres' [I]Topology[/I], 2nd ed. It seems to be the gold standard for intro topology textbooks. It's what my topology course is using, and it's very good and quite readable. Covers all the necessary background in set theory, does the stuff you want for general topology, and even gets into some introductory algebraic topology.
On a related note, I just found out the other day that my topology professor is famous enough to have his own Wikipedia page. (although not a big one)
Thanks for all the help last time, I'm back again with something a little bit more simple.
The issue I am facing right now is not nearly as complicated as the other question I had a few days back.
Do any of you remember grids, with an A and a B, and it asks the number of routes possible to get from A to B with only going S and E, or W and N.
I'm doing that, I have the regular "square grid" problems figured out very easily, but now the grids are not squares, and I am facing some issue. Let me post a photo of an example:
[IMG]http://i.imgur.com/MmIixEH.png[/IMG]
This image is taken right out of my textbook, and I am working on both a and b of question 11.
I've counted the 'spots' on the lines, and hit 13 in total. 7 going west, and 5 going north. This is the case for both examples, so typically, my math would look something like:
[U]13![/U]
7! x 5!
That solution does not give me the correct answer whatsoever, and would not, as it is the same equation for both problems. So, my math is wrong, and I am wondering what exactly I need to change in my problem solving and approach.... Any help is really appreciated!
I don't know any combinatorics so I can't really help you.
For the first one you'll eventually reach a bottleneck where you're at 1 of 2 points so I'm assuming you could take them as two independent results which give the desired answer when multiplied.
For the second one, I think you can take the number of routes possible from A to the top left of the first rectangle (which I suppose you know how to do), and then multiply by the number of routes across the next rectangle (of which there are 2), and then multiply by the number of routes across the last one.
[editline]30th April 2013[/editline]
So you need to acknowledge somehow that there are points removed that make it not rectangular.
[img]https://dl.dropboxusercontent.com/u/4081470/boxes.png[/img]
What Krinkels says is basically correct.
You say you know how to do rectangles, so you need to break up the problem into rectangles (which you know how to do.
For the first one, you either need to go through C [i]or[/i] D, if the path goes through C then it must go through the 2 green rectangles, if it goes through D then it goes through the 2 orange rectangles.
So the number of paths through C is the number of paths through each green rectangle multiplied together.
The number of paths through D is the number of paths through each orange rectangle multiplied together.
Then the total number of paths is just these added together.
The second problem is even simpler, since it must first go through C and then D, so you just multiply the number of paths through each of the three rectangles.
Just finished writing AP Calc (school exam) and it looks like my mark is actually going up! Yay!
Thank you guys a bunch. I really hope you don't mind me coming by and asking for help in math. My teacher rushes through everything, since we do 2 hours of class a day, our school is split up in quarters rather than semesters because it is an Adult 12 school.
If anybody feels like helping me out again, I'm struggling with a bit more...
Basically, it is back to Combinations.
Here is the problem I'm being asked;
How many 5-person committees can be formed from a group of 6 women and 4 men, under each of the following conditions:
a) no conditions
b) must be exactly 3 women
c) must be exactly 4 men
d) there can be no men
e) there must be at least 3 men.
Now, I understand how to solve the a) part of the question, the answer is simple. 10C5 = 252.
Now, I don't know well enough on how to solve the remaining parts of the question.
I know I'm probably annoying, but I really want to learn how to do this stuff so I can pass and get a high mark to enter university. Any help is really appreciated guys, sorry again if I am a pain in the ass.
I think I've got it figured out, and it was really quite simple. I was just making it more difficult than necessary.
To solve b), all I did was calculate 6C3 (6 for total amount of women, 3 for amount I needed), and multiplied it by 4C2 (4 for total amt of men, 2 for amount needed to hit the total number of 5). My answer was 120, which seems to match the answer key in the back of my book.
I haven't heard anything about subsets in class, so I didn't really test whether or not what you said may work for me. It seemed like too much foreign stuff for what I've been learning, though we did do subsets in the unit using venn-diagrams.
What you want to do is the permutations nPr (pick) function for the remaining conditions. The formula we derived in Structures was [img]http://latex.codecogs.com/gif.latex?nPr%20=%20\left%20(%20\frac{n!}{(n-r)!}%20\right%20)[/img]
To do this, n is the group-size, and r is the condition being chosen. (I'm assuming you know the induction principle for factorials to do this where n! = n(n-1)! = n(n-1)(n-2)! etc.)
So in a situation like this, the math must be slightly different in finding the solution, yes?
A youth hostel has 3 rooms that contain 5, 4, and 3 beds, respectively. How many ways can 12 students be assigned to these rooms?
I'm trying to solve it by doing 12C3, (12 = students, 3 = rooms). That obviously isn't going to get me anywhere.
If we assume order of bed assignment matters (i.e. students being assigned to room 1 in this order: 1 2 3 4 5 is distinct from this order: 2 1 3 4 5) then it would be:
[sub]12[/sub]P[sub]5[/sub] + [sub]7[/sub]P[sub]4[/sub] + [sub]3[/sub]P[sub]3[/sub]
Because first there are 12 total students, of which we take 5 in any order. Then there are 7 students left and we take 4 in any order. Finally there are 3 students left and we take them in any order. I think this is correct, might be off though.
And if order of assignment within each room doesn't matter, just use the binomial coefficients instead of permutations. (i.e. [sub]12[/sub]C[sub]5[/sub])
What I know for certain is that the answer to the equation is 27,720.
Doing the math 12P5 + 7P4 + 3P3 = 95,886. So, somewhere along the line, something went wrong.
The math of 12C5 = 792, which is also incorrect.
So, we must be doing something wrong here, or else I am just reading this incorrectly.
Woops, don't know why I added instead of multiplying.
[QUOTE=Yahnich;40494308]it's probably because bed assignment doesn't matter and any student can take any bed given that they don't take so it becomes
12C5 * 7C4 * 3C3
792 * 35 * 1 = 27720[/QUOTE]
Thanks a bunch, man.
Maybe only Joey could answer this, but are Smale's paradox and the Poincare conjecture basically statements about the existence of homotopies, i.e. there is a homotopy between the 2-sphere and the 2-sphere of the opposite orientation and there are homotopies between any closed, simply connected 3-manifold and the 3-sphere?
[editline]1st May 2013[/editline]
Not sure about Poincare, but Smale's paradox definitely seems like it could be phrased as the 2-sphere and the inside-out 2-sphere are homotopic.
We're just starting with topologies.
With sets like {a,b,c} and {pencil,felt pen, marker}
:v:
Another question, I'm sorry. Tell me when to stop and I will.
I've reached another tough point and I am sure I'm just not trying hard enough...
In a standard deck of 52 playing cards, how many different four-card hands are there with one card from each suit?
It's got to be a combination question, not permutation, if my understand is correct.
I tried 52C4. 52 cards, 4 different suits. Answer doesn't meet the correct answer in the key. Sorry to be a pain...
[QUOTE=Yahnich;40495723]there are 13 different cards with the same suit so it's 13C1*4 I think[/QUOTE]
Damn it, I forgot about the 13, which should be common knowledge to me.
The math isn't right though, because the correct answer is apparently 28,561. I've been playing around with my calculator but can't seem to get it no matter what I try.
Sometimes I mess myself up, thinking that there are more parts to the question, and that I need to take nCr and multiply it by another nCr, or something like that. But, it doesn't seem to be the case so far in any of the questions I / we have done tonight.
First, you select your suit for the cards. Since you are getting one of all four, you do 4C1 which = 4. Now you need to select one card of that suit, which is 13C1 = 13. Because you're doing this for all four suits, and you're selecting one card from each, you multiply the two to get the total combinations possible. 13x4=26x2=56 possible outcomes.
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