Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One form of Urysohn's lemma (I suspect there may be more than one) is that on a normal space $(A,\mathcal{T})$ with disjoint closed sets $X$ and $Y$, there exists a continuous real valued function $f$ such that $f|_X=0$ and $f|_Y=1$, and $0\leq f\leq 1$ elsewhere on $A$.

I wanted to know if this could be strengthened to finding a uniformly continuous $f$, but results on Google suggested otherwise.

A hint I found online suggested I find a nonconvergent Cauchy sequence $Z=\{z_n\}$ where $z_m\neq z_n$ when $m\neq n$, and take closed sets $X,Y$ with $X\cap Y=\varnothing$ such that $d(x_n,y_n)\to 0$ for $x_n\in X$, $y_n\in Y$ to be subsequences of that Cauchy sequence.

I asked a previous question looking at $(0,1)$ with the usual metric. User Chris Eagle pointed out that the set $X=\{\frac{1}{2n}\mid n\in\mathbb{N}\}$ and $Y=\{\frac{1}{2n+1}\mid n\in\mathbb{N}\}$ are two such closed sets such that $d(x_n,y_n)\to 0$, that is, $(\lim_{n\to\infty}\vert x_n-y_n\vert=0)$ if we take $x_n=\frac{1}{2n}$ and $y_n=\frac{1}{2n+1}$. I took $z_n=\frac{1}{n}$ to be a Cauchy sequence which doesn't converge in $(0,1)$, and can let $X$ and $Y$ be subsequences of $Z$.

Based on this setup, what fails here so that we can not always expect $f$ to be uniformly continuous?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Let $x_n$ and $y_n$ be nonconvergent Cauchy sequences such that $d(x_n,y_n) \to 0$ as $n\to \infty$. Assume $f$ is a uniformly continuous function such that $f(x_n)=1$ and $f(y_n)=0$. By uniformly continuity, taking $\epsilon = 1/2$, there exists $\delta > 0$ such that whenever $d(u,v) \leq \delta$, we have $|f(u)-f(v)| \leq \epsilon$. Now there exists $N$ such that $d(x_N, y_N) \leq \delta$. But $|f(x_N) - f(y_N)| = |1-0| = 1 > \epsilon$, which obviously contradicts the hypothesis on $f$.

share|improve this answer
    
Thanks Srivatsan! –  Danielle Intal Oct 14 '11 at 3:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.