Recent content by ejionheara

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    Prove: Dense Set A for Continuous f=0 for all x

    Thank you very much for the help!
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    Prove: Dense Set A for Continuous f=0 for all x

    Oh, wow, I'm so used to assuming that going by the definition is wrong. If it's not complicated, it's obviously false... Anyway, now that we've shown that there exists an x in A such |x-a|< delta. Then A intersects this delta neighborhood meaning that f(x) = 0. Then no matter what delta is...
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    Prove: Dense Set A for Continuous f=0 for all x

    Okay, so we proved that there was a contradiction in asserting that f(a) != 0, so f(a) = 0. So then I need to show that there exists an x in A such |x-a| < delta. But A is dense, meaning every open interval has to contain a point of A. Is there a formal way of stating density? Right now it...
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    Prove: Dense Set A for Continuous f=0 for all x

    If I'm correct, if one were to set epsilon = |f(a)|, then wouldn't |f(x)| be strictly between 0 and -2|(f(a)|, (by the reverse triangle inequality). If this is true, then that would be a contradiction because f(x) = 0. Then f(a) = 0, which means a is actually in the set A?
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    Prove: Dense Set A for Continuous f=0 for all x

    No, I do not know those theorems. The class I'm currently in uses Spivak's Calculus (4th) and we just got through the Least Upper Bound chapter. So we really haven't discussed convergence of a sequence at all.
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    Prove: Dense Set A for Continuous f=0 for all x

    Homework Statement A set A of real numbers is said to be dense if every open interval contains a point of A Prove that if f is continuous and f(x) = 0 for all numbers x in a dense set A, then f(x) = 0 for all x. Homework Equations none The Attempt at a Solution The way I understand that...