# Is there a non-compact metric space, every open cover of which has a Lebesgue number?

Lebesgue lemma states that for every open cover $\{U_\alpha\}_{\alpha\in A}$ of a compact metric space $(X,\rho)$ there exists a number $d>0$ such that $$\forall x\in X \quad \exists \alpha_x\in A \quad (r<d \Rightarrow B_r(x)\subset U_{\alpha_x}).$$ I wonder if this property is characteristic for compact metric spaces, that is if for every non-compact metric space there exists an open cover without a Lebesgue number.

Give $\Bbb N$ the discrete metric:

$$d(m,n)=\begin{cases}1,&\text{if }m\ne n\\0,&\text{if }m=n\end{cases}$$

Clearly this space is not compact, but any positive $d\le 1$ is a Lebesgue number for every open cover of it.

Added: Having given the matter a bit more thought, I can prove the following theorem. Say that a metric space $\langle X,d\rangle$ is Lebesgue if every open cover of it has a Lebesgue number.

Theorem: Let $\langle X,d\rangle$ be a metric space. If $X$ has a non-convergent Cauchy sequence or an infinite closed discrete set of non-isolated points, then $X$ is not Lebesgue. In particular, every Lebesgue space is complete, and every perfect Lebesgue space is compact.

Proof: Suppose first that $\sigma=\langle x_k:k\in\omega\rangle$ is a non-convergent Cauchy sequence in $X$. Let $\langle X^*,d^*\rangle$ be the usual metric completion of $\langle X,d\rangle$, and let $p\in X^*$ be the limit of $\sigma$ in $X^*$. Let $V_0=X^*\setminus B_{d^*}(p,2^{-1})$, and for $k>0$ let $V_k=B_{d^*}(p,2^{-k+1})\setminus \operatorname{cl}_{X^*}B_{d^*}(p,2^{-k-1})$. For $k\in\omega$ let $W_k=X\cap V_k$. Then $\mathscr{W}=\{W_k:k\in\omega\}$ is an open cover of $X$ with no Lebesgue number.

Now suppose that $\{x_k:k\in\omega\}$ is a closed discrete set of non-isolated points in $X$. There is a pairwise disjoint, closure-preserving collection $\{V_k:k\in\omega\}$ such that $x_k\in V_k$ for each $k\in\omega$, so there is a sequence $\langle r_k:k\in\omega\rangle$ of positive real numbers such that $B_d(x_k,r_k)\subseteq V_k$ for each $k\in\omega$, and $\langle r_k:k\in\omega\rangle\to 0$. Let $$W=X\setminus\bigcup_{k\in\omega}\operatorname{cl}_X B_d\left(x_k,\frac{r_k}2\right)\;,$$ and let $\mathscr{W}=\{W\}\cup\{B_d(x_k,r_k):k\in\omega\}$; then $\mathscr{W}$ is an open cover of $X$ with no Lebesgue number. $\dashv$

• Thanks. I am asking because as an undergrad I asked one of my teachers whether the property of $X$ "Every continuous function from $X$ is uniformly continuous" is equivalent to $X$ being compact. He didn't know the answer, and again a discrete space was a counterexample. However I recall seeing (much, much later) somewhere that for non-discrete spaces this property is indeed a characterization of compactness for perfect metric spaces. – Asaf Karagila Apr 9 '12 at 9:54
• @Asaf: Sure, take the disjoint union $X = [0,1] \sqcup \mathbb{N}$ where $[0,1]$ and $\mathbb{N}$ where the distance is given by $d(x,y) = |x-y|$ if both $x$ and $y$ are in $[0,1]$, $d(x,y) = 0$ if $x = y \in \mathbb{N}$ and $d(x,y) = 2$ otherwise. – Najib Idrissi Apr 9 '12 at 9:54
• @zulon: Let me rephrase, then. Is there a perfect metric space with the above property which is not compact?. (Because your solution is merely lumping the discrete counterexample onto a compact space. I am actually interested to ask whether all counterexamples are somewhat discrete, or are there counterexamples which look somewhat different) – Asaf Karagila Apr 9 '12 at 9:55
• @Asaf I googled a little and found this: We call Lebesgue spaces the spaces such that every open covering has a Lebesgue number: they turn out to be the spaces X such that every continuous function on X is uniformly continuous. Such spaces are usually called Atsuji spaces. arxiv.org/abs/math/9602203 Maybe some references from this paper will be useful to find out more. – Martin Sleziak Apr 9 '12 at 10:02
• @Asaf: It appears from a quick scan of the paper that Martin found that every Lebesgue metric space is complete and that every perfect Lebesgue metric space is compact. I’ve not thought about the second statement, but I can prove the first one. – Brian M. Scott Apr 9 '12 at 10:16

The property: "every open cover has a lebesgue number" is called Lebesgue property.

For a metric space $M$, having the Lebesgue property is equivalent to:

For any metric spacr $M'$ every continuous function $f:M\rightarrow M'$ is uniforly continuous.

There are other properties equivallent to those, compactness is not one of them.

Here is a counterexample.

Consider the set IN of natural numbers with the the discrete topology, it's not hard to see that IN is not compact and it has the Lebesgue property.