Let c be in S subset R and f: S->R be a function. Assume that for every sequence

Question

Let c be in S subset R and f: S->R be a function. Assume that for every sequence {Xn} subset S with the property that lim Xn=c the sequence {f(Xn)} is convergent. Show that f is continuous at c.

Explanation / Answer

A. this cannot be a vector space because it does not contain the 0-vector. B. we can write 0(x) = 0x^3 + 0x, so the 0-polynomial is an element of S. if p(x) is in S and q(x) in in S, then p(x) = a1x^3 + b1x, q(x) = a2x^3 + b2x therefore, p(x) + q(x) = (a1 + a2)x^3 + (b1 + b2)x and is therefore in S. pick any real number c. if p(x) is in S, then p(x) = ax^3 + bx, so c(p(x)) = c(ax^3 + bx) = (ca)x^3 + (cb)x, so c(p(x)) is in S. S is indeed a subspace. C. the 0-polynomial is not an element of S, since 0(1) = 0 = 0(0). D. here, the 0-vector of R^n is in S, since A(0) = 0. suppose x,y are in S. then A(x+y) = A(x) + A(y) = 0 + 0 = 0, so x+y is in S. let c be any real number, and x in S, so A(x) = 0. then A(cx) = c(A(x)) = c0 = 0, so cx is also in S. S is a subspace. E. the 0-function 0:[a,b]-->R given by 0(x) = 0 for all x in [a,b] is not a member of S. F. the 0-matrix is symmetric. suppose that A,B are symmetric matrices, so that A^T = A and B^T = B. then (A+B)^T = A^T + B^T = A + B, so A+B is symmetric. it is also clear that (cA)^T = c(A^T) = cA, so cA is symmetric as well. this is a subspace. G. the constant 0-function is in S. suppose that f,g are in S. then: (f+g)(0) = f(0) + g(0) = 0 + 0 = 0, so f+g is in S. if c is any real number: (cf)(0) = c(f(0)) = c(0) = 0, so cf is in S if f is. this is also a subspace.

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