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Ε-δ PROOF of ceiling function!

  • Thread starter arpitm08
  • Start date
  • #1
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Homework Statement



Let f: R -> Z be the ceiling function defined by f(x) = ceil(x). Give a ε-δ proof that if a is a real number that is not an integer, then f is continuous at a.


The Attempt at a Solution



I can prove that f(x) is not continuous at any integer. But i don't know how to prove this. I can do proofs for continuous functions, but i've never done one for a piece wise function. Any help would be awesome. Thanks.
 

Answers and Replies

  • #2
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Think about what the graph of this function looks like -- essentially steps that are 1 unit wide. If a is not an integer, it shouldn't be too hard to find a number δ > 0, such that |f(x) - f(a)| < ε. In fact, if x is close enough to a, |f(x) - f(a)| will be 0.
 
  • #3
HallsofIvy
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If a is not an integer, then there exist [itex]\delta> 0[/itex] such that every number in the interval from [itex]a- \delta[/itex] to [itex]a+ \delta[/itex] is not an integer. On that interval f(x) is a constant.
 

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