Steben Posted October 23, 2006 Report Share Posted October 23, 2006 I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics. Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Quote Link to post Share on other sites
Rainman DX Posted October 23, 2006 Report Share Posted October 23, 2006 Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Aaaaahhhhhhhhhh! Make the voices in my head stop! Make the voices go away! Quote Link to post Share on other sites
Bahamut Posted October 23, 2006 Report Share Posted October 23, 2006 I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively. Quote Link to post Share on other sites
Bobwillis Posted October 23, 2006 Report Share Posted October 23, 2006 Anyone else taking the Putnam this year? I'm nervous. It's going to be my first. Quote Link to post Share on other sites
Steben Posted October 23, 2006 Report Share Posted October 23, 2006 I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively. Yeah, H is a subset of the reals. I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me.. Anyone else taking the Putnam this year?I'm nervous. It's going to be my first. I've looked over various Putnam problems (with their solutions handy), but I've never actually taken the test. I should look into it. I probably wouldn't solve any of them, but it'd be fun to try. Quote Link to post Share on other sites
Bahamut Posted October 23, 2006 Report Share Posted October 23, 2006 I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively. Yeah, H is a subset of the reals. I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me.. Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way. Quote Link to post Share on other sites
Steben Posted October 23, 2006 Report Share Posted October 23, 2006 I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively. Yeah, H is a subset of the reals. I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me.. Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way. Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma. Quote Link to post Share on other sites
Bahamut Posted October 23, 2006 Report Share Posted October 23, 2006 I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively. Yeah, H is a subset of the reals. I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me.. Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way. Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma. If the least upper bound and the greatest lower bound isn't in your sets (i. e. if it's an open bounded set), then my method works in constructing a strictly increasing or decreasing sequence. Otherwise, that statement would just be plain false. Quote Link to post Share on other sites
Bobwillis Posted October 23, 2006 Report Share Posted October 23, 2006 I've looked over various Putnam problems (with their solutions handy), but I've never actually taken the test. I should look into it. I probably wouldn't solve any of them, but it'd be fun to try. I wouldn't feel too bad about not solving any of the problems. I've heard that some years the median score is a zero. Quote Link to post Share on other sites
Bahamut Posted October 23, 2006 Report Share Posted October 23, 2006 I've looked over various Putnam problems (with their solutions handy), but I've never actually taken the test. I should look into it. I probably wouldn't solve any of them, but it'd be fun to try. I wouldn't feel too bad about not solving any of the problems. I've heard that some years the median score is a zero. That's usually because many high school students take the test. Quote Link to post Share on other sites
Bobwillis Posted October 23, 2006 Report Share Posted October 23, 2006 That's usually because many high school students take the test. I thought it was for undergraduate mathematics and physics students, only. EDIT: I just read the participation rules. Damn teenagers skewing the results . Quote Link to post Share on other sites
Bahamut Posted October 23, 2006 Report Share Posted October 23, 2006 That's usually because many high school students take the test. I thought it was for undergraduate mathematics and physics students, only. Some advanced public high schools will get their students to take the test I hear. Edit: Yeah...I never actually took the Putnam myself. Some of those tests looked like I could've done well on though. Quote Link to post Share on other sites
Steben Posted October 23, 2006 Report Share Posted October 23, 2006 I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively. Yeah, H is a subset of the reals. I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me.. Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way. Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma. If the least upper bound and the greatest lower bound isn't in your sets (i. e. if it's an open bounded set), then my method works in constructing a strictly increasing or decreasing sequence. That's true. But I need it in general. For instance, the example above. It's not open bounded, so that proof doesn't help me. I've looked over various Putnam problems (with their solutions handy), but I've never actually taken the test. I should look into it. I probably wouldn't solve any of them, but it'd be fun to try. I wouldn't feel too bad about not solving any of the problems. I've heard that some years the median score is a zero. I need to go back and look at some of those problems now that I've taken a math classes where we do actual proofs... maybe I could halfway get one right, if I try real hard. Of course, the real problem with Putnam is that it's on a Saturday during football season. With any luck, I'll be at the SEC tournament that weekend. Quote Link to post Share on other sites
Bahamut Posted October 23, 2006 Report Share Posted October 23, 2006 I've been working on the same stupid theorem in Analysis for the past three weeks. I'm probably just being dumb and missing something obvious, but it's been really difficult... I'll type it below for you guys, at the risk of revealing my relative inexperience at proof-based mathematics.Show that if H is a bounded infinite set, then there exists some sequence made of distinct points of H which is always increasing or always decreasing. Is H a subset of the real numbers? If so, then with a bounded set, you have the least upper bound and greatest lower bound. There must exist sequences in H that converge to the least upper bound and greatest lower bound. Then, there exist subsequences of each sequences such that the subsequences are monotone and converge to the least upper bound and greatest lower bound, respectively. Yeah, H is a subset of the reals. I don't follow your argument, though. Let's say H is the union of {1-1/n|n is a natural number} and {1+1/n|n is a natural number}. It's bounded by 0 and 2, but I need a sequence that is always increasing or always decreasing, rather, x_n>x_n+1 always or x_n<x_n+1 always. No such sequence converges to 0 or 2... the only way I can find such a sequence is if it converges to 1... I can't see how talking about the least upper bound or greatest lower bound helps me.. Yes - I'm just talking about the existence, guaranteed by theorems. If you don't have the particular theorems that I'm referring to, then just prove them - if you have an element that is not the least upper bound or greatest lower bound, you can find another element closer, and keep constructing it that way. Um. I need the sequence to be strictly increasing or strictly decreasing, not monotone. I think that's our delimma. If the least upper bound and the greatest lower bound isn't in your sets (i. e. if it's an open bounded set), then my method works in constructing a strictly increasing or decreasing sequence. That's true. But I need it in general. For instance, the example above. It's not open bounded, so that proof doesn't help me. Oh wait you have an infinite set. The real numbers have the well-ordering property, so you can take that set and find a sequence and order it so that it is always increasing or decreasing then. This argument is quite strong though...(uses the Axiom of Choice) Quote Link to post Share on other sites
Steben Posted October 23, 2006 Report Share Posted October 23, 2006 Oh wait you have an infinite set. The real numbers have the well-ordering property, so you can take that set and find a sequence and order it so that it is always increasing or decreasing then. This argument is quite strong though...(uses the Axiom of Choice) But what if the sequence I chose was {1/4, 1/1, 1/6, 1/1930, 1/205, -4, 1/356!, ...}. I can't order it so that it is always decreasing, because of that -4. Eventually I'll have to use that -4, and then the next number will be greater than it. All that to say, the axiom of choice only lets me pick arbitrary numbers out of the sequence (I think?), and that -4 makes life difficult. That, and, we haven't talked about the Axiom of Choice in class, so I can't use it. Dang. The method I've been attempting to use is to prove that any infinite bounded set has a limit point, and constructing a sequence from that. Unfortunately, I'm having trouble proving any infinite bounded set has a limit point without using the fact that any infinit bounded set has a sequence which is increasing or decreasing. Whoops. Quote Link to post Share on other sites
Luna child Posted October 25, 2006 Report Share Posted October 25, 2006 I only made it to Geometry.. Quote Link to post Share on other sites
Bahamut Posted October 25, 2006 Report Share Posted October 25, 2006 Oh wait you have an infinite set. The real numbers have the well-ordering property, so you can take that set and find a sequence and order it so that it is always increasing or decreasing then. This argument is quite strong though...(uses the Axiom of Choice) But what if the sequence I chose was {1/4, 1/1, 1/6, 1/1930, 1/205, -4, 1/356!, ...}. I can't order it so that it is always decreasing, because of that -4. Eventually I'll have to use that -4, and then the next number will be greater than it. All that to say, the axiom of choice only lets me pick arbitrary numbers out of the sequence (I think?), and that -4 makes life difficult. Edit: Hmm, I'd have to think on it more. That -4 should be irrelevant though. The method I've been attempting to use is to prove that any infinite bounded set has a limit point, and constructing a sequence from that. Unfortunately, I'm having trouble proving any infinite bounded set has a limit point without using the fact that any infinit bounded set has a sequence which is increasing or decreasing. Whoops. Any subset of the real numbers has a limit point - each element is a limit point. That doesn't help you at all though. I'd think more on this, but I got a test tomorrow so maybe tomorrow night I can think on that more. Quote Link to post Share on other sites
Julio Jose Posted October 25, 2006 Report Share Posted October 25, 2006 I've grown to hate math. The day I entered my Abstract Algebra 3 class, and my Linear Algebra 4 class, I came to the conclusion that, holy shit math is a pain in the ass. Only in small doses. He sees why its a pain. Sometimes the methods of proofs just completely elude you even though it looks as though it should be simple. Sometimes it's just a lot of busy work too. It's a little of both. Sometimes in a proof, you will spend so much time working backwards to move forwards, it's not even funny. Other times, it's because there is just a ridiculus amount of work required by these classes. Assignments really do take 2 weeks at a time. You can't just bs around and do them 1 or 2 days before they are do. It takes a lot of commitment, and as someone who isn't even majoring in the pmat field, it's kind of crazy. Quote Link to post Share on other sites
Steben Posted October 25, 2006 Report Share Posted October 25, 2006 The method I've been attempting to use is to prove that any infinite bounded set has a limit point, and constructing a sequence from that. Unfortunately, I'm having trouble proving any infinite bounded set has a limit point without using the fact that any infinit bounded set has a sequence which is increasing or decreasing. Whoops. Any subset of the real numbers has a limit point - each element is a limit point. That doesn't help you at all though. I'd think more on this, but I got a test tomorrow so maybe tomorrow night I can think on that more. The definition of a limit point we're using is that "x is a limit point of a set H iff any segment about x contains an element of H distinct from x". Any subset of the real numbers contains a limit point of the real numbers, but I need a limit point of the set itself. Quote Link to post Share on other sites
Steben Posted October 26, 2006 Report Share Posted October 26, 2006 Here's some calculus, because I know it'll make your day. An engineering friend asked me if this problem was possible, I said I didn't think so, but I'm also incredibly dumb, so maybe you can? Here's what he asked. Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative? Quote Link to post Share on other sites
Sahaquiel Delta Posted October 26, 2006 Report Share Posted October 26, 2006 Bleah. I'm only in Calculus I. (For the second time. I hate the public high school system. I mean, I got a damn A the first time...) All you advanced college peoples make me sad (that I'm not in Calc II and such). Quote Link to post Share on other sites
Ab56 v2 aka Ash Posted October 27, 2006 Report Share Posted October 27, 2006 I don't think calc 2 would teach you how to integrate that function; it doesn't look like integration by parts would simplify it. I'm intrigued though Quote Link to post Share on other sites
Jabe Posted October 27, 2006 Report Share Posted October 27, 2006 Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative? I don't think so because u is a generic function. Without any specifics, you can't determine the output of int[(u')^2]dx without resorting to just evaluating on a case-to-case basis. For example, there is no way to explain the following: If u(x)=x then the function (int[(u')^2]dx) = x+C or possibly u'(x)*u(x) If u(x)=ln(x) then the function = -1/x+C or possibly -u'(x) Not really a proof, but it's what I say. Quote Link to post Share on other sites
Steben Posted October 27, 2006 Report Share Posted October 27, 2006 Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative? I don't think so because u is a generic function. Without any specifics, you can't determine the output of int[(u')^2]dx without resorting to just evaluating on a case-to-case basis. For example, there is no way to explain the following: If u(x)=x then the function (int[(u')^2]dx) = x+C or possibly u'(x)*u(x) If u(x)=ln(x) then the function = -1/x+C or possibly -u'(x) Not really a proof, but it's what I say. That's the reasoning I used. I don't think calc 2 would teach you how to integrate that function; it doesn't look like integration by parts would simplify it. I'm intrigued though It doesn't. Here's what happened when I tried integration by parts. int((u')^2)dx = int(u'u')dx = uu' - int(u''u)dx = uu' - (uu' - int(u'u')dx) = int(u'u')dx = int((u')^2)dx Quote Link to post Share on other sites
Bahamut Posted October 27, 2006 Report Share Posted October 27, 2006 Here's some calculus, because I know it'll make your day. An engineering friend asked me if this problem was possible, I said I didn't think so, but I'm also incredibly dumb, so maybe you can? Here's what he asked.Is it possible to simplify int[(u')^2]dx to an expression that doesn't use an integral symbol, where u is a function of x and u' is its first derivative? Think of differential equations. In this case, the equation would be f' = (u')^2. I can't think of any generalized solution to this, can you? Quote Link to post Share on other sites
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