Answer to Question #160018 in Real Analysis for Luna

Question

A function is uniformly continuous on a set I⊂R if

a) for some ε₀>0 we can find a δ>0 such that if x,y∈I and |x-y|<δ, then |f(x)-f(y)|<ε₀

b) for any ε₀>0 we can find a δ>0 such that if x,y∈I and |x-y|<δ, then |f(x)-f(y)|<ε₀

c) for any ε₀>0 and any δ>0, if x,y∈I and |x-y|<δ₀, then f(x)-f(y)|<ε₀

d) all of the above

e) none of the above

Answer 1

$\displaystyle \textbf{\textsf{Option B is correct}}\\ \textsf{Given a function}\,\, f: D \to R \\ \textsf{as above and an element}\,\, x_0 \\ \textsf{of the domain}\,\, D,\,\, f \,\, \textsf{is said}\\ \textsf{to be continuous at the point}\,\,x_0 \\ \textsf{when the following holds:}\\ \textsf{For any number}\,\, \epsilon > 0, \,\, \textsf{however small,}\\ \textsf{there exists some number}\,\, \delta > 0\\ \textsf{such that for all}\,\, x \,\, \textsf{in the domain}\\ \textsf{of}\,\, f \,\, \textsf{with}\,\, x_0 − \delta < x < x_0 + \delta, \,\, \textsf{the value of}\,\, f(x) \,\, \textsf{satisfies}\\ f(x_{0})-\varepsilon <f(x)<f(x_{0})+\varepsilon.\\ \textsf{Alternatively written, continuity of}\,\, f : D \to R \\ \textsf{at}\,\, x_0 \in D\,\, \textsf{means that for every}\,\, \epsilon > 0 \\ \textsf{there exists a}\,\, \delta > 0 \\ \textsf{such that for all}\,\, x \in D: |x-x_{0}|<\delta \Rightarrow |f(x)-f(x_{0})|<\varepsilon$