# 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.

#### Solutions Collecting From Web of "Is there a non-compact metric space, every open cover of which has 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. $\dashv$

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.