# clarification errata in Munkres Topology?

While reading the second edition of Munkres' Topology, I came across this (page 129):

Theorem 21.1 Let $f: X \rightarrow Y$; let $X$ and $Y$ be metrizable with metrics $d_X$ and $d_Y$, respectively. Then continuity of $f$ is equivalent to the requirement that given $x \in X$ and given $\epsilon > 0$, there exists $\delta > 0$ such that $d_X(x,y) \implies d_Y(f(x), f(y)) < \epsilon$.

Shouldn't the last part be $d_X(x,y) < \delta \implies d_Y(f(x), f(y)) < \epsilon$ ? I've looked at the errata here but didn't find mention of this. am i missing something? Thanks for any help/clarification. :)

• You are correct: it’s missing a $<\delta$. – Brian M. Scott Sep 6 '12 at 23:20
• You're right. In my 7th printing of the second edition this is corrected: see here. – t.b. Sep 6 '12 at 23:20
• I wanted to post a one word answer, but there is a minimum verbosity required :-(. – copper.hat Sep 6 '12 at 23:39
• @copper.hat: You can use ${}{}{}...{}$ or $\phantom{filler}$ or something to that effect :) (works in comments, too) – t.b. Sep 6 '12 at 23:42
• @t.b.: ${}$T${}$h${}$a${}$n${}$k${}$s! – copper.hat Sep 7 '12 at 0:10

To have an answer: Yes, you are right, there's a $\lt \delta$ missing here.
Theorem 21.1. $\ \$ Let $f:X\to Y$; let $X$ and $Y$ be metrizable with metrics $d_X$ and $d_Y$, respectively. Then continuity of $f$ is equivalent to the requirement that given $x\in X$ and given $\epsilon>0$, there exists $\delta>0$ such that $$d_X(x,y)<\delta\implies d_Y(f(x),f(y))<\epsilon.$$