A challenge for those who have seen a bit of analysis before.
A map \(f\colon X\to Y\) is continuous (everywhere) if
\[ \forall x\in X\, \forall \epsilon>0\, \exists \delta>0\, \forall x’\in X\, d(x,x’)<\delta \Longrightarrow d(f(x),f(x’))<\epsilon. \]
On the other hand, it is uniformly continuous if
\[\forall \epsilon>0\, \exists \delta>0\, \forall x\in X\, \forall x’\in X\, d(x,x’)<\delta \Longrightarrow d(f(x),f(x’))<\epsilon. \]
Check your understanding of quantifier-ology by convincing yourself that uniformly continuous implies continuous, but not conversely (e.g. the map \(x\mapsto x^2\) from the real line to itself is continuous but not uniformly continuous).
The argument we gave for the continuity of the Peano curve definitely does not prove uniform continuity, because (as we said in class), for fixed \(x\), the number \(\delta\) such that if \(|x-x’|<\delta\) then \(x\) and \(x’\) agree up to \(m\) digits in their ternary expansion depends on the ternary expansion of \(x\). On the other hand, it is a general theorem that when \(X\) is compact, e.g. the unit interval, every continuous map actually is uniformly continuous. Explain this apparent inconsistency.