Let (X,d1) and (Y,d2) two metric spaces and let f:X-Y be an onto continuous function s.t d1(x,y)<=d2(f(x),f(y)) for all in X. Prove if (X,d1) is complete then (Y,d2) is also complete
131535
Easy Homework.\n\nSuppose {y_n} is a Cauchy sequence in Y. Since f is onto, there are x_n with f(x_n)=y_n. Since d1(x_n,x_m)<=d2(y_n,y_m), {x_n} is Cauchy. Since X is complete, there is a limit, x of {x_n}. By continuity, {y_n} converges to f(x). Thus every Cauchy sequence in Y converges, so Y is complete.
664555
Say the point (1,2) is on f(x) and the point (2,5) is on g(x).\nNow since (1,2) is on f that means f(1) = 2.\nSimilarly, g(2) = 5 since (2,5) is on g.\nYou can now find (gof))(1)\n(gof)(1) = g[f(1)]  = g{2) = 5\n(1,5) is a point on (gof)(x)\nThe y-value of one function has to match the x-value of the other one.\nLet (4,7) be on f and (3,4) be on g.  You can find (fog)(3)\n(fog)(3) = f(g(3)) = f(4) = 7\n(3,7) is a point on (fog)(x)
30450
The question is actually asking you if cosine(x) is a continuous function. if a function is continuous, lim(f(x)) as x approach a is f(a).\n\na function is continuos if the following criteria are satisfied:\n1. the function f(x) exists for all x\n2. for all x, lim(f(x)) from the positive side = lim from the negative side for all points.\n\n\nnow since cosine(x) is a function that satisfies both criteria, cos(x) is continuous and therefore lim(cos(x)) as x approaches a is just cos(a).
393069
1. f(g(x))\n2. g(f(x))
418905
HOMEWORK\nTo find f, just collapse A. To find g, make sure the image is inside of A.\nSeparability is easy: in fact, there is a dense *point*. To answer about pseudometrizability, figure out what balls would have to look like.\n\nYou really should do your own work for math at this level. It is no help  to your understanding if you just ask others. If you really need help, ask your professor.
475636
If a=2 & b=3 then f(x) is continuous for any x but x=1, because when x-->1 from left f(x)=|x-1|=0 & when x-->1 from right f(x)=a*x^2+b*x=2+3=5, 0 is not = 5.\nFunction f(x) is continuous if f(x)[from right]=f(x)[from left], that is a*1^2+b*1=0, or b=-a\nThus for any a & b=-a the whole f(x) is continuous.\nf(x) would have derivative for any x if f’(x)[from left]=f’(x)[from right] & f(x) [from left]=f(x)[from right] for any x. The point x=1 is particular node. f’(x)[from left]=-1, then 2*a*1+b*1 must also be = -1, or b=2*a-1 & on the other hand b must be = -a, hence a=1/3 & b=-1/3; thus the only case when the whole f(x) is continuous with its derivative exists at a=1/3 & b=-1/3
7110
(Y-y1) = m(X-x1)\non point (4,7) => Y-7 = m(X-4)  (I)\non point (2,5) => Y-5 = m(X-2)  (II)\n\n(I&II) ==> m = 1
414878
1 -> b\n2 -> e\n3 -> c\n4 -> d\n5 -> a\n6 -> f
442107
You have made an error in the description of the problem, because inf  A is not the lowest upper bound of A because that is Sup A.\n\nGiven: two non empty sets of real numbers;\nA , B   ; For all a in A & for all b in B  : a ≤ b  (####)\n\na#). . .if for all a ε A & a2 ε A :  a ≤  a2 => a2 = Sup A \nb#). . .if for all b ε B & b1 ε B:   b1 ≤  b => b1 = Inf B\nNow:\na#) & b#) =>  a2 ≤ b1 <=> because of (####) <=> Sup A ≤ Inf B\n\nProof of question 1):\nThere exist exactly one number c ε R : b ≤ c ≤ a for all a ε A & b ε B\n(@@@)  b ≤ c ≤ a  & Sup a ≤ Inf B =>\n Inf B  ≤ Sup A & Sup A ≤ Inf B =>\nSup A = Inf B  put this into (@@@) => \nInf B ≤ c ≤  Sup A & Sup A ≤ c ≤ inf B => \nc = Sup A = Inf B (%%%)\n\n2) Is already in proved look (%%%)\n\n3) I suppose you mean b in B instead t in T. \na  ≤  b =>  0  ≤  b-a   this will always be true because Sup A =  Inf B, but I can't say it is true for all z in Z+ because, I don't now anything about Sup B , but if it has now upper limit it will be true for all z in Z+
641073
Try to graph the points and then decide what the best function that can fit the point; however , I tried the following:\n\nSince a, b, c, d are constants then \n\ny=(a-d)/(1+(x/c)^b)+d \n\nLet a-b = k and c^b = L then\n\ny= k /(1+(x^b)/L) + d \n\ny= (k * L)/(L+(x^b)) + d\n\nLet k*L = m\n\ny= m / (L+(x^b) + d\n\ny= ( m + L*d + d*x^b ) / ( L + x^b)\n\nLet u= m+L*d\n\ny= (  u + d*x^b ) / ( L + x^b)\n\nI think the graph of such a function as smiler to the (ln) function since both the nominator and the denominator have the same order. But I couldn't simplify the function any more..... that might help a bit ...! good luck
388269
F(x) is the same as y. You don't really solve this, its the graph of a line. You can graph it if thats what asked and then one of the points in the graph would be the ordered pair that you are looking for. To get an ordered pair, just plug in a number, any number, for x OR y. Then solve for the other variable. The original number you put in and the resultant number make the ordered pair.
471288
I'm assuming A is not empty. Then, by the definition of the topology T, it follows A is in T, so that A is open and intB = B. Right.\n\nclB = B? No! B is a subset of clB and since B is not empty, for it contains non-empty A, it follows clB is not empty. Wrong.\n\nNow, see. In the topology T, a set S is closed if it's complement is open. Therefore, S is closed if  S = X or if the complement of S contains A, which, in this case, implies S doesn't intersect A,. clB is the intersection f all closed sets containing B. So, the only closed set that contains B is X, so that clB = X.\n\nWe have that Fr(B) = clB - intB = X - B, so that Fr(B) is the complement of B. therefore, unless B = X, your guess is wrong.\n\nJust check, I may have made a mistake.
442668
uniform continuity implies continuity, but not conversely.  The first link seems to help quit a bit, the wikipedia not so much so, though it does have a good example with the details missing.\n\n\nhttp://www.maths.nott.ac.uk/personal/jff/G12RAN/pdf/Uniform.pdf#search='uniform%20continuity'\n\nhttp://en.wikipedia.org/wiki/Uniform_continuity
508231
1. If you pass a vertical line down the graph, and it only passes through the function once, then ya, it is function.  If it passes through more than once, then it is not a function.\n\n2. Isolate x
430449
It's not right.\n\nf: G -> G so that f(x)=1 for every x in G is a homomorphism but it\nisn't a homomorphism.\n\nIf f isn't trivial you can prove the statement.\nf homomorphism => f(G) and Kerf are subgroups of G.\nG of prime order implies (according to Lagrange's theorem) that its only subgroups are {1} and G itself. \nIf f isn't trivial Kerf /= G (not equal) so Kerf={1} and f is injective.\nOn the other side f not trivial implies f(G) /= {1} therefore we have\nf(G)=G and f is surjective. So f is a isomorphism.\n\nExcuse me, but my English sucks :-)
481339
The first condition for a function to be differentiable in a point x is that the function is continuous at that point.\nit's easy to prove the function is continuous at x=1\nbecause:\nLim f(x)=Lim(x^2 + 1)= 2 for x going to 1 from left (1-)\nLim f(x)=Lim(2x)=2 for x going to 1 from right (1+)\nWe have now proved that the function is continuous at x=1\n\nWe will look at f' at x=1\nf'(x)=2x ; x<=1\nf'(x)=2 ; x>1\nthe function is differentiable in a point x=a if and only if, the function is continuous and f' exist and has the same value when x goes to point x=a from left and right\n\nLim f'(x)=2 for x going to x=1 from left (1-)\nLim f'(x)=2 for x going to x=1 from right (1+)\nnow we have proven that the function is differentiable at x=1
271109
Okay, let's take this one part at a time:\n\n"If the function is not continuous at c, then in many cases (as with rational functions) the function is continuous all around it, but there is a discontinuity at that isolated point."\n\nConsider the function f(x)=x²/x. As long as x!=0, this function evaluates to a definite value, and is continuous (basically the line y=x). However, at x=0, this funtion yeilds 0/0, which is formally undefined, and since the function doesn't have a value at that point, it cannot be continuous there. This is the sort of function the text is talking about.\n\n"In that case, we want to find a similar function, except with the hole filled in."\n\nWe want to find a continuous function which is the same as the first function everywhere the first function is defined. A nice example of such a function for our example would be f(x)=x, which gives the same result as the first function everywhere the first function is defined, but has the advantage of being defined at x=0.\n\n"The limit of this function at c will be the same, as can be seen from the definition of a limit."\n\nThe limit of this function will be the same, as the original text will shortly do a poor job of explaining.\n\n"The function is the same as the previous except at a point c."\n\nThe function is the same as the previous except at a point c.\n\n"The limit definition depends on f(x) only at the points where 0 < |x - c| < δ."\n\nThis is a reference to the delta-epsilon definition of a limit. This guy - δ - is the greek small letter delta. The basic idea of a limit is that as you get closer and closer to the point you are evaluating the limit at, the values will get closer and closer to the limit itself, and so if you got arbitrarily close to the point you are evaluating the limit at, your values would get arbitrarily close to the limit. This is formalized in the delta-epsilon definition by saying that "(x→c)lim f(x)=L iff for all ε>0, there exists δ>0 such that for all x where 0<|x-c|<δ, |f(x)-L|<ε" or in plainer language: "the limit of a function at c is L iff no matter how small a positive nonzero number ε you name, you can name a second positive nonzero number δ such that all points whose horizontal distance from c is between zero and δ (exclusive), the difference between the value of the function at these points and L will always be less than ε." And yes, the delta-epsilon definition is hard to understand, even in plain language, but it's basically just a formalization of the intuitive description I provided.\n\n"When x = c, that inequality is false, and so the limit at c does not depend on the value of the function at c."\n\nThe limit of a function at c only depends on what happens to it as you approach c, not what happens to it when you finally get there.\n\n"Therefore, the limit is the same."\n\nIt doesn't matter that our new function differs from the old one at point c, because it's the same everywhere else, and it's only what happens everywhere else that affects the limit. Therefore, the limit is the same.\n\n"And since our new function is continuous, we can now just evaluate the function at c as before."\n\nSince our new function is continuous, we can figure out is limit at c by simply evaluating it at c, which would then in turn give us the limit of our old function at c (since these limits are the same). For instance, we can figure out the limit of f(x)=x²/x is at zero by reducing it to f(x)=x and evaluating that at zero, giving us (unsurprisingly) zero.\n\nAnd by the way, whoever wrote the textbook you pulled that definition out of needs to be fired. The only way you could possibly understand that explanation is if you already knew the concept that the explanation was trying to explain, thus rendering the explanation redundant.