Let $X$ be a metric space, and let $a$ and $b$ be points in $X$ such that $a\ne b$. By the first axiom of metric spaces, this means that $d(a,b)>0$. We can choose $r$ with $$ and put $U=OB(a,r)$ and $V=OB(b,r)$. Then $U$ and $V$ are open, and $a\in U$ and $b\in V$, and we have $U\cap V=\mathrm{\varnothing}$. This is an obvious and natural construction that occurs frequently in the theory of metric spaces.
Now suppose that $X$ is a general topological space, and we again have points $a,b\in X$ with $a\ne b$. We might again want to have open sets $U$ and $V$ with properties as above, but there is no longer any obvious way to produce them, and in fact, they need not exist in all cases. This leads us to introduce the following definition.
Let $X$ be a topological space. If $a,b\in X$ with $a\ne b$, then a Hausdorff separation for $a$ and $b$ is a pair of open sets $U$ and $V$ such that $a\in U$ and $b\in V$ and $U\cap V=\mathrm{\varnothing}$. We say that $X$ is a Hausdorff space if every pair of distinct points has a Hausdorff separation.
As in our previous discussion, if $a$ and $b$ are distinct points in a metric space $X$, then the open balls $U=OB(a,d(a,b)/2)$ and $V=OB(b,d(a,b)/2)$ give a Hausdorff separation of $a$ and $b$. Thus, all metric spaces are Hausdorff spaces.
Let $X$ be a space with the indiscrete topology, as in Example 3.15, so the only open sets are $\mathrm{\varnothing}$ and $X$. Suppose also that $|X|\ge 2$, so we can choose two distinct points $a,b\in X$ with $a\ne b$. It is then clear that there can be no Hausdorff separation of $a$ and $b$ (because both sets $U$ and $V$ would have to be equal to $X$). Thus, $X$ is not Hausdorff.
The Sierpiński space $X$ is defined as follows: the underlying set is $\{0,1\}$, and the sets $\mathrm{\varnothing},\{1\}$ and $\{0,1\}$ are declared to be open, but the set $\{0\}$ is not open. One can check that this is indeed a topology, and that there is no Hausdorff separation for $0$ and $1$, so we do not have a Hausdorff space.
Consider the line with doubled origin $X/E$ as in Example 7.27. This contains two distinct points $a=\pi (0,1)$ and $b=\pi (0,-1)$. One can check that there is no Hausdorff separation for these points, so $X/E$ is not a Hausdorff space. In particular, this illustrates the fact that a quotient of a Hausdorff space need not be Hausdorff.
Although non-Hausdorff spaces are important in various different branches of mathematics, we will mostly restrict attention to Hausdorff spaces.
Let $X$ be a Hausdorff space, and let $Y$ be a subset of $X$, with the subspace topology. Then $Y$ is also Hausdorff.
Let $a$ and $b$ be distinct points in $Y$. As $X$ is Hausdorff, we can choose sets $U,V$ that are open in $X$ with $a\in U$ and $b\in V$ and $U\cap V=\mathrm{\varnothing}$. Put ${U}^{\prime}=U\cap Y$ and ${V}^{\prime}=V\cap Y$. By the definition of the subspace topology, these are open in $Y$. It is also clear that $a\in {U}^{\prime}$ and $b\in {V}^{\prime}$ and ${U}^{\prime}\cap {V}^{\prime}=U\cap V\cap Y=\mathrm{\varnothing}$, so ${U}^{\prime}$ and ${V}^{\prime}$ give a Hausdorff separation of $a$ and $b$ in $Y$.
∎
Let $Y$ and $Z$ be disjoint topological spaces, and put $X\mathrm{=}Y\mathrm{\cup}Z$, and give $X$ the coproduct topology as in Definition 7.1. Suppose that $Y$ and $Z$ are both Hausdorff; then $X$ is also Hausdorff.
Consider a pair of distinct points $a,b\in X$. There are four possible cases:
Both $a$ and $b$ lie in $Y$
Both $a$ and $b$ lie in $Z$
$a$ is in $Y$ and $b$ is in $Z$
$a$ is in $Z$ and $b$ is in $Y$.
In case (a), we use the fact that $Y$ is Hausdorff. We can thus find sets $U$ and $V$ that are open in $Y$ with $a\in U$ and $b\in V$ and $U\cap V=\mathrm{\varnothing}$. From the definition of the coproduct topology we see that $U$ and $V$ are still open when considered as subsets of $X$, so they provide the required Hausdorff separation of $a$ and $b$. Case (2) is essentially the same. In case (3) the pair $(Y,Z)$ is the required Hausdorff separation, and in case (4) we use $(Z,Y)$ instead.
∎
Let $Y$ and $Z$ be topological spaces, and consider the space $Y\mathrm{\times}Z$ with the product topology. Suppose that $Y$ and $Z$ are both Hausdorff; then $Y\mathrm{\times}Z$ is also Hausdorff.
Consider a pair of distinct points $a=(y,z)\in Y\times Z$ and $b=(u,v)\in Y\times Z$. As $a\ne b$, we must either have $y\ne u$ or $z\ne v$. First suppose that $y\ne u$ in $Y$. As $Y$ is Hausdorff, we can choose a Hausdorff separation $(U,V)$ for $y$ and $u$ in $Y$. It is then not hard to see that the pair $(U\times Z,V\times Z)$ is a Hausdorff separation for $a$ and $b$ in $Y\times Z$. Similarly, if $z\ne v$ then we can choose a Hausdorff separation $(P,Q)$ for $z$ and $v$ in $Z$, and we find that the pair $(Y\times P,Y\times Q)$ is a Hausdorff separation for $a$ and $b$ in $Y\times Z$.
∎
We now turn to the notion of compactness. Compactness for metric spaces was covered in some detail in [MS, Section 7]. In particular, it was shown as [MS, Theorem 7.8] that a subset $X\subseteq {\mathbb{R}}^{n}$ is compact iff it is bounded and closed. For our purposes, you should think of this as a prototypical example: a space is compact if it is similar to a bounded, closed subset of ${\mathbb{R}}^{n}$. It was also proved in [MS, Theorem 7.20] that a metric space is compact iff it has the Heine-Borel property. In the context of general topological spaces, we take the Heine-Borel property as the definition of compactness. We now recall the relevant details.
Let $X$ be a topological space. By an open cover of $X$ we mean a family of open subsets ${({U}_{i})}_{i\in I}$ such that $X={\bigcup}_{i\in I}{U}_{i}$. Equivalently, for every point $x\in X$ there must exist a set ${U}_{i}$ in the family such that $x\in {U}_{i}$. By a finite subcover we mean a subcollection ${U}_{{i}_{1}},\mathrm{\dots},{U}_{{i}_{r}}$, containing only finitely many of the subsets ${U}_{i}$, such that we still have $X={U}_{{i}_{1}}\cup \mathrm{\cdots}\cup {U}_{{i}_{r}}$. We say that $X$ is compact if every open cover of $X$ has a finite subcover.
Consider the space $\mathbb{R}$. The sets ${U}_{n}=(n-1,n+1)$ (for $n\in \mathbb{Z}$) form an open cover of $\mathbb{R}$. However, we claim that there is no finite subcover. In other words, if we take any finite collection of these sets, say ${U}_{{n}_{1}},\mathrm{\dots},{U}_{{n}_{r}}$, we claim that the union ${U}^{*}={U}_{{n}_{1}}\cup \mathrm{\cdots}\cup {U}_{{n}_{r}}$ is not all of $\mathbb{R}$. This is clear: if $p=\mathrm{min}({n}_{1},\mathrm{\dots},{n}_{r})$ and $q=\mathrm{max}({n}_{1},\mathrm{\dots},{n}_{r})$ then ${U}^{*}\subseteq (p-1,q+1)$ and so $q+2\notin {U}^{*}$, for example. From this it follows that $\mathbb{R}$ is not compact. (We could also have deduced this from the results in [MS, Section 7].)
Now consider the space $X=(0,1)$ and the subspaces ${U}_{n}=(1/n,1-1/n)$ (for $n>1$). These form an open cover of $X$ but there is no finite subcover, so $X$ is again not compact.
Let $X$ be a subset of ${\mathbb{R}}^{n}$. We say that $X$ is bounded if there is a constant $R\ge 0$ such that $\parallel x\parallel \le R$ for all $x\in X$.
Let $X$ be a space with only finitely many points; then $X$ is compact.
Supose that $X=\{{x}_{1},\mathrm{\dots},{x}_{n}\}$. Let ${({U}_{i})}_{i\in I}$ be an open cover of $X$. For each $t$, the element ${x}_{t}$ must lie in some set of the cover, say ${x}_{t}\in {U}_{{i}_{t}}$. This means that $X\subseteq {U}_{{i}_{1}}\cup \mathrm{\cdots}\cup {U}_{{i}_{n}}$, so we have a finite subcover as required. ∎
Let $X$ be a subset of ${\mathbb{R}}^{n}$, with the subspace topology. Then $X$ is compact iff it is bounded, and closed in ${\mathbb{R}}^{n}$.
This follows from [MS, Theorems 7.8 and 7.20]. (It is apparently identical to Theorem 7.8, but that is slightly misleading because [MS] uses a different definition of compactness; we need Theorem 7.20 as well to show that the two versions of compactness are the same.) ∎
The spaces ${S}^{n-1}$, ${\mathrm{\Delta}}^{n-1}$ and ${B}^{n}$ are bounded and closed in ${\mathbb{R}}^{n}$, so they are all compact.
Recall that ${O}_{n}=\{A\in {M}_{n}(\mathbb{R})|{A}^{T}A=I\}$. We will show that this is closed and bounded in ${M}_{n}(\mathbb{R})\simeq {\mathbb{R}}^{{n}^{2}}$, so it is compact. We can define $f:{M}_{n}(\mathbb{R})\to {M}_{n}(\mathbb{R})$ by $f(A)={A}^{T}A-I$. This is continuous, and $\{0\}$ is closed in ${M}_{n}(\mathbb{R})$, and ${O}_{n}={f}^{-1}\{0\}$, so ${O}_{n}$ is closed in ${M}_{n}(\mathbb{R})\simeq {\mathbb{R}}^{{n}^{2}}$. We have also explained that the standard metric on ${M}_{n}(\mathbb{R})$ can be expressed in the form
$$d(A,B)=\sqrt{\text{trace}({(A-B)}^{T}(A-B))}.$$ |
For the identity matrix $I\in {M}_{n}(\mathbb{R})$ we have $\text{trace}(I)=n$. From this we see that $d(A,0)=\sqrt{n}$ for all $A\in {O}_{n}$, so ${O}_{n}$ is bounded.
Let $X$ be a compact space, and let $Y$ be a closed subset of $X$, equipped with the subspace topology. Then $Y$ is also compact.
Let ${({V}_{i})}_{i\in I}$ be an open cover of $Y$. By the definition of the subspace topology, we can find sets ${U}_{i}$ that are open in $X$ such that ${V}_{i}={U}_{i}\cap Y$. As $Y={\bigcup}_{i}{V}_{i}$, we see that $Y\subseteq {\bigcup}_{i}{U}_{i}$. Note that the set ${Y}^{c}=X\setminus Y$ is also open in $X$ (because $Y$ is assumed to be closed). The set ${Y}^{c}$ together with the sets ${U}_{i}$ then form an open cover of $X$. As $X$ is compact, we can choose a finite subcover, which will consist of some finite collection of sets ${U}_{{i}_{1}},\mathrm{\dots},{U}_{{i}_{r}}$, possibly together with ${Y}^{c}$. These sets cover all of $X$, so in particular they cover $Y$. It is clear that ${Y}^{c}$ cannot contribute to covering $Y$, so we must have $Y\subseteq {U}_{{i}_{1}}\cup \mathrm{\cdots}\cup {U}_{{i}_{r}}$. This means that $Y={V}_{{i}_{1}}\cup \mathrm{\cdots}\cup {V}_{{i}_{r}}$, so we have a finite subcover of the original cover, as required. ∎
Let $X$ be a Hausdorff space, and let $Y$ be a subset that is compact with respect to the subspace topology. Then $Y$ is closed in $X$.
We must show that the set ${Y}^{c}=X\setminus Y$ is open. Choose a point $x\in {Y}^{c}$. For each point $y\in Y$ we have $x\ne y$, so we can choose a Hausdorff separation, say $({U}_{y},{V}_{y})$. This means that ${U}_{y}$ and ${V}_{y}$ are open in $X$ and $x\in {U}_{y}$ and $y\in {V}_{y}$ and ${U}_{y}\cap {V}_{y}=\mathrm{\varnothing}$. Now put ${W}_{y}={V}_{y}\cap Y$, which is open for the subspace topology on $Y$. As $y\in {W}_{y}$, we see that the sets ${W}_{y}$ cover $Y$. By the compactness property, we can choose a finite subcollection ${W}_{{y}_{1}},\mathrm{\dots},{W}_{{y}_{r}}$ that still covers $Y$. This means that $Y\subseteq {V}_{{y}_{1}}\cup \mathrm{\cdots}\cup {V}_{{y}_{r}}$. Now put ${U}^{\mathrm{\#}}={U}_{{y}_{1}}\cap \mathrm{\cdots}\cap {U}_{{y}_{r}}$. This is a finite intersection of open sets, so it is still open in $X$. Each of the sets ${U}_{{y}_{i}}$ contains $x$, so we have $x\in {U}^{\mathrm{\#}}$. We also claim that ${U}^{\mathrm{\#}}\subseteq {Y}^{c}$, or equivalently ${U}^{\mathrm{\#}}\cap Y=\mathrm{\varnothing}$. To see this, suppose (for a contradiction) that we have $y\in {U}^{\mathrm{\#}}\cap Y$. This means that $y\in Y$, and $Y\subseteq {V}_{{y}_{1}}\cup \mathrm{\cdots}\cup {V}_{{y}_{r}}$, so we have $y\in {V}_{{y}_{t}}$ for some $t$. On the other hand, we have $y\in {U}^{\mathrm{\#}}={U}_{{y}_{1}}\cap \mathrm{\cdots}\cap {U}_{{i}_{r}}$, so in particular $y\in {U}_{{y}_{t}}$. We now see that $y\in {U}_{{y}_{t}}\cap {V}_{{y}_{t}}$, which is impossible because ${U}_{z}\cap {V}_{z}=\mathrm{\varnothing}$ for all $z$. This contradiction shows that we in fact have ${U}^{\mathrm{\#}}\subseteq {Y}^{c}$. We started with an arbitrary element $x\in {Y}^{c}$, and we produced an open set ${U}^{\mathrm{\#}}$ that contains $x$ and is contained in ${Y}^{c}$. By Lemma 3.33, this is enough to prove that ${Y}^{c}$ is open. ∎
Let $X$ be a compact Hausdorff space, and let $Y$ be a subset of $X$. Then $Y$ is closed in $X$ iff it is compact (in the subspace topology).
Let $X$ and $Y$ be topological spaces. Suppose that $X$ is compact, and that we have a surjective, continuous map $f\mathrm{:}X\mathrm{\to}Y$. Then $Y$ is also compact.
Consider an open cover ${({V}_{i})}_{i\in I}$ of $Y$. Put ${U}_{i}={f}^{-1}({V}_{i})\subseteq X$, so that ${U}_{i}$ is open by the definition of continuity. If $x$ is any point in $X$, then $f(x)\in Y$, and the sets ${V}_{i}$ cover $Y$, so we can choose $i$ with $f(x)\in {V}_{i}$, which means that $x\in {U}_{i}$. This shows that the sets ${U}_{i}$ form an open cover of $X$. As $X$ is assumed to be compact, there must be a finite subcover, say ${U}_{{i}_{1}},\mathrm{\dots},{U}_{{i}_{r}}$. We claim that the corresponding sets ${V}_{{i}_{1}},\mathrm{\dots},{V}_{{i}_{r}}$ cover $Y$. To see this, consider an arbitrary point $y\in Y$. As $f$ is assumed to be surjective, we can choose $x\in X$ with $f(x)=y$. As the sets ${U}_{{i}_{1}},\mathrm{\dots},{U}_{{i}_{r}}$ cover $X$, we can choose $t$ such that $x\in {U}_{{i}_{t}}$. Here ${U}_{{i}_{t}}$ was defined to be ${f}^{-1}({V}_{{i}_{t}})$, so we must have $f(x)\in {V}_{{i}_{t}}$, or in other words $y\in {V}_{{i}_{t}}$ as required. Thus, the list ${V}_{{i}_{1}},\mathrm{\dots},{V}_{{i}_{r}}$ is a finite subcover of our original cover. ∎
Let $X$ be a compact space, and let $E$ be an equivalence relation on $X$. Then the quotient space $X\mathrm{/}E$ is also compact.
The quotient map $\pi :X\to X/E$ is surjective and continuous, so this follows from Proposition 8.20. ∎
We can define an equivalence relation $E$ on ${S}^{n}$ by $xEy$ iff $y=\pm x$. The real projective space $\mathbb{R}{P}^{n}$ is defined to be the quotient space ${S}^{n}/E$. This is compact, because ${S}^{n}$ is compact. Similarly, we can regard ${S}^{2n+1}$ as the unit sphere in the space ${\u2102}^{n+1}={\mathbb{R}}^{2n+2}$. We can then define another equivalence relation $F$ on ${S}^{2n+1}$ by declaring that $xFy$ iff $y={e}^{i\theta}x$ for some $\theta \in \mathbb{R}$. The complex projective space $\u2102{P}^{n}$ is defined to be ${S}^{2n+1}/F$; this is again compact.
Let $Y$ and $Z$ be disjoint spaces, and take $X\mathrm{=}Y\mathrm{\cup}Z$, with the coproduct topology. If $Y$ and $Z$ are both compact, then so is $X$.
Let ${({U}_{i})}_{i\in I}$ be an open cover of $X$. Then each set ${U}_{i}$ must have the form ${V}_{i}\cup {W}_{i}$, where ${V}_{i}$ is an open subset of $Y$, and ${W}_{i}$ is an open subset of $Z$. It is easy to see that the sets ${V}_{i}$ form an open cover of $Y$, so there must be a finite subcover, say ${Y}_{{i}_{1}},\mathrm{\dots},{Y}_{{i}_{r}}$. Similarly, the sets ${W}_{i}$ form an open cover of $Z$, so there must be a finite subcover, say ${W}_{{j}_{1}},\mathrm{\dots},{W}_{{j}_{s}}$. We then find that the list ${U}_{{i}_{1}},\mathrm{\dots},{U}_{{i}_{r}},{U}_{{j}_{1}},\mathrm{\dots},{U}_{{j}_{s}}$ forms a finite subcover of our original cover. ∎
Let $Y$ and $Z$ be compact spaces. Then the product $Y\mathrm{\times}Z$ is also compact (under the product topology).
We will give the proof after some preliminary definitions and results.
Let $Y$ and $Z$ be as above, and let ${({U}_{i})}_{i\in I}$ be an open cover of $Y\times Z$. We will say that a subset $K\subseteq Y\times Z$ is finitely covered if there is a finite subcollection ${U}_{{i}_{1}},\mathrm{\dots},{U}_{{i}_{r}}$ such that $K\subseteq {U}_{{i}_{1}}\cup \mathrm{\cdots}\cup {U}_{{i}_{r}}$.
We need to prove that the whole space $Y\times Z$ is finitely covered, but we will build up to this by proving that certain other sets are finitely covered first.
In the above context, let $y$ be a point in $Y$. Then there is an open set $S\mathrm{\subseteq}Y$ such that $y\mathrm{\in}S$ and $S\mathrm{\times}Z$ is finitely covered.
For each $z\in Z$, choose an index ${i}_{z}$ such that $(y,z)\in {U}_{{i}_{z}}$. As ${U}_{{i}_{z}}$ is open in the product topology, we can then choose a box ${B}_{z}={V}_{z}\times {W}_{z}$ such that $y\in {V}_{z}$ and $z\in {W}_{z}$ and ${V}_{z}\times {W}_{z}\subseteq {U}_{{i}_{z}}$. As $z\in {W}_{z}$ for all $z$, we see that the sets ${({W}_{z})}_{z\in Z}$ form an open cover of $Z$. As $Z$ is compact, we can choose a finite subcover, say $Z={W}_{{z}_{1}}\cup \mathrm{\dots}\cup {W}_{{z}_{r}}$. Now put $S={V}_{{z}_{1}}\cap \mathrm{\cdots}\cap {V}_{{z}_{r}}$. This is the intersection of a finite list of open sets, each of which contains $y$; so $S$ itself is an open set containing $y$. We next claim that $S\times Z\subseteq {B}_{{z}_{1}}\cup \mathrm{\cdots}\cup {B}_{{z}_{r}}$. Indeed, suppose we have a point $(s,z)\in S\times Z$. As $Z={W}_{{z}_{1}}\cup \mathrm{\dots}\cup {W}_{{z}_{r}}$, we can choose $k$ such that $z\in {W}_{{z}_{k}}$. Now $s\in S={V}_{{z}_{1}}\cap \mathrm{\cdots}\cap {V}_{{z}_{r}}$, so we also have $s\in {V}_{{z}_{k}}$. This proves that $(s,z)\in {V}_{{z}_{k}}\times {W}_{{z}_{k}}={B}_{{z}_{k}}$, as required. We also have ${B}_{{z}_{k}}\subseteq {U}_{{i}_{{z}_{k}}}$, so
$$S\times Z\subseteq {U}_{{i}_{{z}_{1}}}\cup \mathrm{\cdots}\cup {U}_{{i}_{{z}_{r}}}.$$ |
This shows that $S\times Z$ is finitely covered. ∎
For each $y\in Y$, the lemma tells us that we can choose an open set ${S}_{y}$ containing $y$ such that ${S}_{y}\times Z$ is finitely covered. The family ${({S}_{y})}_{y\in Y}$ is then an open cover of $Y$. As $Y$ is assumed to be compact we can choose a finite subcover, say $Y={S}_{{y}_{1}}\cup \mathrm{\cdots}\cup {S}_{{y}_{r}}$. It follows that $Y\times Z={\bigcup}_{i=1}^{r}({S}_{{y}_{i}}\times Z)$. Here each of the subsets ${S}_{{y}_{i}}\times Z$ is finitely covered, and there are only finitely many of them, so the union is also finitely covered. This means that $Y\times Z$ is finitely covered, or in other words that our original cover of $Y\times Z$ has a finite subcover, as required. ∎
Let $X$ be a compact space, and let $Y$ be a Hausdorff space. Let $f\mathrm{:}X\mathrm{\to}Y$ be a continuous bijection. Then the inverse map ${f}^{\mathrm{-}\mathrm{1}}\mathrm{:}Y\mathrm{\to}X$ is also continuous, so $f$ is a homeomorphism.
Let $g:Y\to X$ be the inverse of $f$, so the claim is that $g$ is continuous. By Proposition 3.23, it will be enough to check that for every closed subset $F\subseteq X$, the preimage ${g}^{-1}(F)$ is closed in $Y$. Here $F$ is a closed subset of a compact space, so it is compact by Proposition 8.17. As $g$ is inverse to $f$, we have ${g}^{-1}(F)=f(F)$. We can regard $f$ as a continuous surjective map from $F$ to $f(F)$ (where $F$ and $f(F)$ are given the respective subspace topologies). It follows by Proposition 8.20 that $f(F)$ is also compact. Now $f(F)$ is a compact subspace of the Hausdorff space $Y$, so it is closed by Proposition 8.18. In other words, the preimage ${g}^{-1}(F)$ is closed, as required. ∎
In Example 7.24 we considered a space $X$ consisting of two disjoint discs, and the space $Y={S}^{2}$. We introduced an equivalence relation $E$ on $X$ (corresponding to the idea of gluing the boundary circles of the two discs) and defined a continuous bijection $\overline{f}:X/E\to Y$. Now $X/E$ is a quotient of a closed bounded subset of ${\mathbb{R}}^{3}$, so it is compact, and $Y$ is a metric space, so it is Hausdorff. It follows from Proposition 8.27 that $\overline{f}$ is actually a homeomorphism.
In Example 7.26 we constructed a continuous bijection $\overline{f}:{\mathbb{R}}^{2}/E\to T$, where $T$ is the torus and the equivalence relation $E$ is given by $xEy$ iff $x-y\in {\mathbb{Z}}^{2}$. Here $T$ is a metric space and therefore Hausdorff. It is easy to see that the composite
$${[0,1]}^{2}\stackrel{\text{inc}}{\to}{\mathbb{R}}^{2}\stackrel{\mathit{\pi}}{\to}{\mathbb{R}}^{2}/E$$ |
is surjective, and ${[0,1]}^{2}$ is compact, so ${\mathbb{R}}^{2}/E$ is compact by Proposition 8.20.
We conclude with some slightly different results about open covers, that are only relevant for metric spaces. Their use will become apparent later.
Let $X$ be a metric space, and let ${({U}_{i})}_{i\in I}$ be an open cover. A Lebesgue number for this cover is a number $\u03f5>0$ with the following property: for every point $x\in X$, there is an index $i$ such that $OB(x,\u03f5)\subseteq {U}_{i}$.
You should think of a Lebesgue number as something like the minimum size of overlaps between adjacent sets in the cover.
Let $X$ be a compact metric space. Then every open cover of $X$ has a Lebesgue number.
Let ${({U}_{i})}_{i\in I}$ be an open cover of $X$. For each point $x\in X$, we can choose an index ${i}_{x}$ such that $x\in {U}_{{i}_{x}}$. As ${U}_{{i}_{x}}$ is open, we can then choose ${r}_{x}>0$ such that $OB(x,{r}_{x})\subseteq {U}_{{i}_{x}}$. Put ${V}_{x}=OB(x,{r}_{x}/2)$, so ${V}_{x}$ is an open subset of $X$ containing $x$. The sets ${V}_{x}$ then form an open cover of $X$, so we can choose a finite subcover, say $X={V}_{{x}_{1}}\cup \mathrm{\cdots}\cup {V}_{{x}_{n}}$. Put $\u03f5=\mathrm{min}({r}_{{x}_{1}},\mathrm{\dots},{r}_{{x}_{n}})/2$. We claim that this is a Lebesgue number. To see this, let $x$ be an arbitrary point in $X$. As the sets ${V}_{{x}_{1}},\mathrm{\dots},{V}_{{x}_{n}}$ form a cover, we can choose $k$ such that $x\in {V}_{{x}_{k}}$, or in other words $$. If $u\in OB(x,\u03f5)$ then $$ and $$ so $$. This proves that $OB(x,\u03f5)\subseteq OB({x}_{k},{r}_{{r}_{k}})\subseteq {U}_{{i}_{{x}_{k}}}$, so $OB(x,\u03f5)$ is contained in one of the sets ${U}_{i}$, as required. ∎