How can I prove the continuity of the ceiling function at a non-integer value?

  • Thread starter Thread starter arpitm08
  • Start date Start date
  • Tags Tags
    Function Proof
Click For Summary
SUMMARY

The discussion focuses on proving the continuity of the ceiling function, defined as f(x) = ceil(x), at non-integer values. It is established that f(x) is continuous at any real number a that is not an integer, as there exists a δ > 0 such that for all x within the interval (a - δ, a + δ), f(x) remains constant and equal to f(a). The proof hinges on the observation that within this interval, the ceiling function does not change its value, thus satisfying the ε-δ definition of continuity.

PREREQUISITES
  • Understanding of the ε-δ definition of continuity
  • Familiarity with piecewise functions
  • Basic knowledge of the ceiling function and its properties
  • Graphical interpretation of step functions
NEXT STEPS
  • Study the ε-δ definition of continuity in depth
  • Explore proofs involving piecewise functions
  • Analyze the properties of the ceiling function in various contexts
  • Learn about discontinuities in functions, specifically at integer values
USEFUL FOR

Mathematics students, educators, and anyone interested in real analysis, particularly those studying continuity and piecewise functions.

arpitm08
Messages
50
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
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.
 

Similar threads

Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
4K
Replies
8
Views
2K