shrinking lemma (changes) in nLab

Proof

Since U 1∪U 2=XU_1 \cup U_2 = X it follows (by de Morgan's law) that their complements X\U iX \backslash U_i are disjoint closed subsets. Hence by normality of (X,τ)(X,\tau) there exist disjoint open subsets

V 1⊃X\U 2AAAV 2⊃X\U 1. V_1 \supset X \backslash U_2 \phantom{AAA} V_2 \supset X \backslash U_1 \,.

By their disjointness, we have the following inclusions:

V 1⊂X\V 2⊂U 1. V_1 \subset X \backslash V_2 \subset U_1 \,.

In particular, since X\V 2X \backslash V_2 is closed, this means that Cl(V 1)⊂X\V 2Cl(V_1) \subset X \backslash V_2.

Hence it only remains to observe that V 1∪U 2=XV_1 \cup U_2 = X, by definition of V 1V_1.

Proof

By induction using lemma 2.

To begin with, consider {U 1,∪i=2nU i}\{ U_1, \underoverset{i = 2}{n}{\cup} U_i\}. This is a binary open cover, and hence lemma 2 gives an open subset V 1⊂XV_1 \subset X with V 1⊂Cl(V 1)⊂U 1V_1 \subset Cl(V_1) \subset U_1 such that {V 1,∪i=2nU i}\{V_1, \underoverset{i = 2}{n}{\cup} U_i\} is still an open cover, and accordingly so is

{V 1}∪{U i} i∈{2,⋯,n}. \{ V_1 \} \cup \left\{ U_i \right\}_{i \in \{2, \cdots, n\}} \,.

Similarly we next find an open subset V 2⊂XV_2 \subset X with V 2⊂Cl(V 2)⊂U 2V_2 \subset Cl(V_2) \subset U_2 and such that

{V 1,,V 2}∪{U i} i∈{3,⋯,n} \{ V_1, ,V_2 \} \cup \left\{ U_i \right\}_{i \in \{3, \cdots, n\}}

is an open cover. After nn such steps we are left with an open cover {V i⊂X} i∈{1,⋯,n}\{V_i \subset X\}_{i \in \{1, \cdots, n\}} as required.

Proof

As in the proof of lemma 3, there exist V iV_i for i∈ℕi \in \mathbb{N} such that V i⊂Cl(V i)⊂U iV_i \subset Cl(V_i) \subset U_i and such that for every finite number, hence every n∈ℕn \in \mathbb{N}, then

∪i=0nV i∪∪i=n+1∞U i=∪i=0nXU i. \underoverset{i = 0}{n}{\cup} V_i \cup \;=\; \underoverset{i = 0}{n}{\cup} n+1}{\infty}{\cup} U_i \,. \;=\; X\,.

Now the extra assumption that {U i⊂X} i∈I\{U_i \subset X\}_{i \in I} is locally finite implies that every x∈Xx \in X is contained in only finitely many of the U iU_i, hence that for every x∈Xx \in X there exists n x∈ℕn_x \in \mathbb{N} such that

x ∈ ∉∪i=0n x+1n x∞U i. x \in \notin \underoverset{i = 0}{n_x}{\cup} n_x+1}{\infty}{\cup} U_i \,.

This implies that for every eachxx then

x∈∪i=0n xV i⊂∪i∈ℕV i x \in \underoverset{i = 0}{n_x}{\cup} V_i \subset \underset{i \in \mathbb{N}}{\cup} V_i

hence that {V i⊂X} i∈ℕ\{V_i \subset X\}_{i \in \mathbb{N}} is indeed a cover of XX.

Proof

of the general shrinking lemma 1

Let {U i⊂X} i∈I\{U_i \subset X\}_{i \in I} be the given locally finite cover of the normal space (X,τ)(X,\tau). Consider the set SS of pairs (J,𝒱)(J, \mathcal{V}) consisting of

1. a subset J⊂IJ \subset I;

2. an II-indexed set of open subsets 𝒱={V i⊂X} i∈I\mathcal{V} = \{V_i \subset X\}_{i \in I}

with the property that

1. (i∈J⊂I)⇒(Cl(V i)⊂U i)(i \in J \subset I) \Rightarrow ( Cl(V_i) \subset U_i );

2. (i∈I\J)⇒(V i=U i)(i \in I \backslash J) \Rightarrow ( V_i = U_i ).

3. {V i⊂X} i∈I\{V_i \subset X\}_{i \in I} is an open cover of XX.

Equip the set SS with a partial order by setting

((J 1,𝒱)≤(J 2,𝒲))⇔((J 1⊂J 2)and(∀i∈J 1(V i=W i))). \left( (J_1, \mathcal{V}) \leq (J_2, \mathcal{W}) \right) \Leftrightarrow \left( \left( J_1 \subset J_2 \right) \,\text{and}\, \left( \underset{i \in J_1}{\forall} \left( V_i = W_i \right) \right) \right) \,.

By definition, an element of SS with J=IJ = I is an open cover of the required form.

We claim now that a maximal element (J,𝒱)(J, \mathcal{V}) of (S,≤)(S,\leq) has J=IJ = I.

For assume on the contrary that there were i∈I\Ji \in I \backslash J. Then we could apply the construction in lemma 2 to replace that single V iV_i with a smaller open subset V′ iV'_i to obtain 𝒱′\mathcal{V}' such that Cl(V′ i)⊂V iCl(V'_i) \subset V_i and such that 𝒱′\mathcal{V}' is still an open cover. But that would mean that (J,𝒱)<(J∪{i},𝒱′)(J,\mathcal{V}) \lt (J \cup \{i\}, \mathcal{V}'), contradicting the assumption that (J,𝒱)(J,\mathcal{V}) is maximal. This proves by contradiction that a maximal element of (S,≤)(S,\leq) has J=IJ = I and hence is an open cover as required.

We are reduced now to showing that a maximal element of (S,≤)(S,\leq) exists. To achieve this we invoke Zorn's lemma. Hence we have to check that every chain in (S,≤)(S,\leq), hence every totally ordered subset has an upper bound.

So let T⊂ST \subset S be a totally ordered subset. Consider the union of all the index sets appearing in the pairs in this subset:

K≔∪(J,𝒱)∈TJ. K \;\coloneqq\; \underset{(J,\mathcal{V}) \in T }{\cup} J \,.

Now define open subsets W iW_i for i∈Ki \in K picking any (J,𝒱)(J,\mathcal{V}) in TT with i∈Ji \in J and setting

W i≔V iAAAi∈K. W_i \coloneqq V_i \phantom{AAA} i \in K \,.

This is independent of the choice of (J,𝒱)(J,\mathcal{V}), hence well defined, by the assumption that (T,≤)(T,\leq) is totally ordered.

Moreover, for i∈I\Ki \in I\backslash K define

W i≔U iAAAi∈I\K. W_i \coloneqq U_i \phantom{AAA} i \in I \backslash K \,.

We claim now that {W i⊂X} i∈I\{W_i \subset X\}_{i \in I} thus defined is a cover of XX. Take an arbitrary point x∈Xx \in X. If x∈U ix \in U_i for some i∉Ki \notin K, we have U i=W iU_i = W_i and therefore xx is in ∪i∈IW i\underset{i \in I}{\cup} W_i. Otherwise, combining with the assumption that {U i⊂X} i∈I\{U_i \subset X\}_{i \in I} is locally finite, the set J xJ_x of indices i∈Ii \in I such that x∈U ix \in U_i is finite and J x⊂KJ_x \subset K. Since (T,≤)(T,\leq) is a total order, it must contain an element (J,𝒱)(J, \mathcal{V}) such that J x⊂JJ_x \subset J. And since that 𝒱\mathcal{V} is a cover and xx cannot belong to any U iU_i with ii outside of J xJ_x, it must be that x∈∪i∈J xV i⊂∪i∈JV ix \in \underset{i \in J_x}{\cup} V_i \subset \underset{i \in J}{\cup} V_i, and hence xx is in ∪i∈IW i\underset{i \in I}{\cup} W_i.

This shows that (K,𝒲)(K,\mathcal{W}) is indeed an element of SS. It is clear by construction that it is an upper bound for (T,≤)(T ,\leq ). Hence we have shown that every chain in (S,≤)(S,\leq) has an upper bound, and so Zorn’s lemma implies the claim.