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Let $(X,d)$ be a metric space and $(a_n)$ be a sequence of distinct points in $ X$ such that each $a_n$ is a limit point of $X$. If $U_n$ 's are mutually disjoint open neighbourhoods of $a_n$ in $X$. Then the function $f:X→R$ given by $f(x)=\frac{d(x,A_n)}{n(d(x,A_n)+d(x,b_n))}$ if $x∈\overline{U_n}$ for some $n$ ;$f(x)=0$ otherwise where $A_n={a_n}∪(\overline{U_n}−U_n)$ and $b_n$ is point in $U_n$ different from $a_n$.

S. Nadler in his paper POINTWISE PRODUCTS OF UNIFORMLY CONTINUOUS FUNCTIONS SARAJEVO JOURNAL OF MATHEMATICS Vol.1 (13) (2005), 117–127 (Theorem 4.2) constructed this function and said that it is a uniformly continuous function, which I am not able to verify. I think without some extra conditions on the sets Un's it wont be possible to prove the uniform continuity of this function (as we do not have nice pasting lemma for uniformly continuous functions as for continuous fucntions). Please comment.(Actually, the paper does not show explicitly that the function is uniformly continuous. see the last line on page 121' it follows easily that f is uniformly continuous on all of X.' but it does not seem to be easy with the information provided by the author of the paper.I am having problem in finding suitable delta for overlapping parts of $\overline{U_n}$ and $\overline{U_m}$)

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Maybe the closures of $U_n$ are also disjoint? If not, how would you define $f$ on the intersection of closures? Note that making them disjoint is trivial, just diminish their radius twice. – Yulia Kuznetsova Jan 16 '12 at 16:09
@Yulia: on the boundary of each $U_n$, $f(x) = 0$ in each $\overline{U_n}$ by definition, so $f$ is well-defined. – Jaap Eldering Jan 16 '12 at 16:12

I think one can prove this along the following lines.

First, if we take a finite subset $[0,M] \subset \mathbb{N}$ and restrict to those $n \in M$, then clearly $f$ is uniformly continuous.

Secondly, $0 \le \|f(x)\| \le \frac{1}{n}$ on $U_n$.

Let $\epsilon > 0$. Choose $M$ large enough that $\frac{2}{M} < \epsilon$. Now if we have two points $x_1, x_2 \in X$, then if they both do not lie within any $U_n$ for $n > M$, then use the uniform continuity of the subset $[0,M]$. If both $x_i \in U_k$ for some $k > M$, then $\|f(x_1) - f(x_2)\| < \epsilon$ by the second, direct estimate on $f$.

If one $x$ lies inside $U_k$ for some $k > M$ and the other doesn't, then use the triangle inequality with a third point $y \in \overline{U_k} - U_k$ and the above estimates again.

I think that with some careful examination, this can be turned into a rigorously defined uniform continuity modulus for $f$.

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