Diffrentiation implies continuiy but do continuity implies differntiation????
599545
No. Look at f(x) = |x|. It is everywhere continuous but not differentiable with respect to x at x=0.\n\nThis is a stupid question, and you're stupid for asking it.
77303
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).
928313
When descriptive containment is equivalent to topological containment:\n\nhttp://megafoundation.org/CTMU/Articles/IntroCTMU.htm
492808
The statement that no such space exists is called the Continuum Hypothesis. It was hypothesized by Cantor, but he gave no proof of it. Godel showed that it is consistent with the usual axioms of set theory to assume CH. Then Paul Cohen showed that it is also consistent with set theory to assume CH is false. In other words, the Continuum Hypothesis is independent of the axioms of set theory. You can either assume it or assume its negation without introducing inconsistencies.\n\nThat said, there are mathematicians that are trying to find other axioms that would either imply CH or its negation. These other axioms would have to have some independent reason for including them in the system, of course. Typically, their attempts have to do with the existence of 'large cardinals'.
899585
There is continuity and the preservation of tradition despite change.
910126
I'm assuming this is with metric spaces D with the metric d and ℝ with the Euclidean metric.\n\nProof.\n\nLet ε > 0. Since both f and g are uniformly continuous, ∃ δ₁, δ₂ > 0 such that for ∀ x, y ∈ D,\n\nd(x, y) < δ₁ ⇒ |f(x) - f(y)| < ε / 2, and\nd(x, y) < δ₂ ⇒ |g(x) - g(y)| < ε / 2.\n\nLet δ = min(δ₁, δ₂). Then d(x, y) < δ implies\n\n|(f+g)(x) - (f+g)(y)|\n= |f(x) + g(x) - f(y) + g(y)|\n= |f(x) - f(y) + g(x) - g(y)|\n≤ |f(x) - f(y)| + |g(x) - g(y)|\n< ε / 2 + ε / 2\n= ε. ∎
74348
If you look at the standard epsilon/delta proof that G(x) = x^2 is continuous 0, it all works for F(x) too. \n\nThat is, given epsilon (e), you can find delta (d) such that /G(x) - G(0)/ < e when /x - 0/ < d.\n(Since for -1 < x < 1,  x^2 < |x|, just pick d = min(e, 1))\n\nNow looking at F(x), given d, pick that same e you would use with G(x). If x is rational, F(x) = G(x)\nso /G(x) - G(0)/ = /F(x) - F(0)/ < e.  If x is irrational, /G(x) - G(0)/ = /0 - 0/ < e.
436747
What we learn from the book A Brief History of Time is that time and space as spatial parts of the cordinate system can't be separated from one another in a model of an ever expandng Universe. We know, or have a idea, that before the Universe started expanding it had no space cordinates .Hence as time and space cordinates are inseparable it is only safe to assume that time had a start just as the universe was created. \n\nSo the answer to the question is No.
212931
Impulsive power is not constant, it is sporadic and irregular.  Continuous power is constant and regular.
432891
If g(x) is differentiable, then it certainly is continuous, and hence integrable.  You can define H(x) as the following definite integral:\n\nH(x) = 1 + Integral from 0 to x of { g(t) dt }
324083
This is an easy problem.  Here are the rough ingredients:\n\n(1) f is an isometry (i.e. d(f(x),f(y)) = d(x,y) for all x,y in X) implies that f is continuous.  (Prove this!)\n\n(2) The image of a compact set under a continuous map is a compact set, and so f(X) is a compact subset of X.\n\n(3) Suppose, for contradiction, that f is not onto.  So there is x in X &#92; f(X).  Let epsilon be positive real number less than dist(x,f(X)) (this distance is defined to be the infimum of the distances d(x,y) for all y in f(X)).  Since X is compact, X can be cover by finitely many balls of radius epsilon.  Let N be the smallest number of balls of radius epsilon with which you are able to cover X.  x belongs to one of these balls, obviously, and such a ball is disjoint from f(X).  So there is a cover of f(X) consisting of at most N-1 balls of radius epsilon.  But the inverse images (under f) of these balls (all of them of radius epsilon because f is an isometry) cover X.  Since we chose N to be minimal, and we have found now an epsilon-cover of X consisting of N-1 balls, we have reached a contradiction.
615364
Uniform composition; continuity
966571
A layer of seperation between "things".
472839
No.\n\nDifference between my answer and the answer above?  Above = short term perspective.  Below = long term perspective.
737727
What does C represent?  If C is a constant, then this doesn't converge.\nWhat is the context?  What is the name of the class, chapter and section that this comes from?
589780
The definition of continuity that the preimage of any open set is open, is elegant but not directly amenable to intuition in my opinion. Maybe an easier way of putting it (logically equivalent) is the following. It's much closer to the the usually concept of continuity but applies to any topological spaces:\n\nA function f:X-->Y is continuous if and only if it preserves limit point relationships, that is: If A is any subset of X and p a limit point of A, then f(p) is a limit point, or point, of f(A).\n\nIf every open set S containing f(p) has an open preimage f^(-1)(S), then p being the limit point of some set A means that points of A are contained in f^(-1)(S) --> points of f(A) are contained in S. The converse is also easily proved so this is an equivalent definition of continuity.\n\nThe concept of a limit point p of a set A is relatively easy to grasp. It just means there's no open neighborhood of p which doesn't contain points of A - there's no finite margin separating p from points of A.\n\nIf f is continuous, for any point p we can typically construct a sequence A having p as its unique limit point, then the definition of continuity tells us that the sequence of image points under f is f(p).
887496
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