let $X = (0,1)$ and $U_n = (\frac{1}{n},1)$ then find the Lebesgue number of the set . I know the meaning of Lebesgue number.Now, I have to find a number $a$ such that if $A \subset X$ and $diam(A) < a$ then $A$ will be contained in $U_i$. The metric is not defined so I dont know how to proceed. I know that diam($A$)= $a$ means that $\sup\{d(x,y)\}=a$ where $x,y \in A$ . Something which is confusing me is that will the set $(0,a)$ and $(a,2a)$ where $a<\frac{1}{2}$ have the same diameter . If so then it is not possible to cover them using the same cover I guess..
 A: Usually, Lebesgue number is defined for coverings of compact spaces only.
It would be $0$ here; to see this, take any $\delta>0$ and $n > 1/\delta$ and then take a point set $\{\frac{1}{n}, \frac{1}{n+1}, \ldots\}$ which has diameter less then $\delta$ and is not contained in a single of these sets.
A: The answer depends on the metric. So let us assume that the metric on $(0,1)$ is the "usual" metric ($d(x,y) = \lvert  x  - y \rvert$). Let $\lambda > 0$  and $r = \min(1, \lambda/2)$. Then $(0,r)$ is a subset of $(0,1)$ with diameter $< \lambda$, but it is not contained in any $U_n$. Thus no $\lambda > 0$ can be a Lebesgue number.
If we use another metric, then the answer may be different. Here is an example. The map
$$h :(0,1) \to (1,\infty), h(x) = 1/x$$
is a homeomorphism. Thus the usual metric on $(1,\infty)$ induces a metric on $(0,1)$ which is given by $d(x,y) = \lvert h(x) - h(y) \rvert$. Then any $\lambda > 0$ is a Lebesgue number (or, if want to express it that way, $\infty$ is a Lebesgue number). Let $A \subset (0,1)$ be a set which is not contained in any $U_n$. Then we can find points $x_n \in A$ such that $x_n \notin U_n$, i.e. $´0 < x_n \le 1/n$. But then for all $n$
$$\text{diam}(A) \ge d(x_n,x_1) = \lvert 1/x_n - 1/x_1 \rvert \ge 1/x_n - 1/x_1 \ge n - 1/x_1$$
which shows that $\text{diam}(A) = \infty$.
