6

Given a normal space $\Omega$.

Then closed sets can be separated continuously: $$h\in\mathcal{C}(\Omega,\mathbb{R}):\quad h(A)\equiv0,\,h(B)\equiv1\quad(A,B\in\mathcal{T}^\complement)$$ Especially, it can be chosen as a bump: $0\leq h\leq1$

Though the idea is very clear it can be strikingly technical.

The goal here is to crystalize out the most lucid version!

Disclaimer: This is a record for future threads.

C-star-W-star
  • 16,275

3 Answers3

6

$\newcommand{\cl}{\operatorname{cl}}$I prefer a more topological proof.

Let $D$ be the set of dyadic rationals in $[0,1]$. We shall construct a family $\{U(d):d\in D\}$ of sets $A \subseteq U(d) \subseteq \Omega\setminus B$, all open with the possible exception of $U(0)$, satisfying the condition that $\cl U(d_0)\subseteq U(d_1)$ whenever $d_0,d_1\in D$ and $d_0<d_1$.

For $n\in\Bbb Z^+_0$ = set of non-negative integers let $D_n = \{\frac{k}{2^n} : k\in\{0,\ldots,2^n\} \}$. We have $D_{n} \subseteq D_{n+1}$ and $\bigcup_{n=0}^\infty D_n = D$. We shall construct the desired $U(d)$ inductively for $d \in D_n$.

For $n = 0$ let $U(0)=A$ and $U(1)=\Omega\setminus B$. Note that $U(0)$ is closed, but in general not open.

Now suppose the $U(d)$ have been constructed for $d \in D_n$. By normality, for each $k\in\{0,\ldots,2^n\}$ there is an open $U\left(\frac{2k+1}{2^{n+1}}\right)$ such that

$$\cl U\left(\frac{k}{2^n}\right)=\cl U\left(\frac{2k}{2^{n+1}}\right)\subseteq U\left(\frac{2k+1}{2^{n+1}}\right)\subseteq\cl U\left(\frac{2k+1}{2^{n+1}}\right)\subseteq U\left(\frac{k+1}{2^n}\right)\;.$$

Let

$$h:\Omega\to[0,1]:x\mapsto\begin{cases} 1,&\text{if }x\notin U(1)\\ \inf\{d\in D:x\in U(d)\},&\text{otherwise}\;. \end{cases}$$

Clearly $h[A]=\{0\}$ and $h[B]=\{1\}$, so it only remains to show that $h$ is continuous. First note that if $r\in D$ and $x\in U(r)$, then $h(x)\le r$, while if $x\in\Omega\setminus\cl U(r)$ then $h(x)\ge r$. Now let $x\in\Omega$ and $\epsilon>0$ be arbitrary.

  • If $h(x)=0$, choose $r\in D$ so that $0<r<\epsilon$; then $V=U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq[0,\epsilon)$.

  • If $h(x)=1$, choose $r\in D$ so that $1-\epsilon<r<1$; then $V=\Omega\setminus\cl U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq(1-\epsilon,1]$.

  • If $0<h(x)<1$, choose $r,s\in D$ so that $h(x)-\epsilon<r<h(x)<s<h(x)+\epsilon$; then $V=U(s)\setminus\cl U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq(h(x)-\epsilon,h(x)+\epsilon)$.

Thus, $h$ is continuous at every point of $\Omega$.

Paul Frost
  • 76,394
  • 12
  • 43
  • 125
Brian M. Scott
  • 616,228
  • Can you explain the main difference of both constructions? (It seems to me they're basically the same, or?) – C-star-W-star Jan 16 '15 at 18:18
  • @Freeze_S: The underlying idea is the same; the details and point of view are not. Rudin’s version is, unsurprisingly, the sort of argument that an analyst would naturally make; the other version avoids semicontinuity altogether. Rudin also leaves quite a bit of detail in the construction of the $V_s$ for the reader to fill in. – Brian M. Scott Jan 16 '15 at 18:23
  • You claim "by induction one can define inductively" while this proof is actually constructive (but only does it for a sufficient subset of the rationals); it's the proof I have been taught as well. It's very direct, no extra notions necessary. – Henno Brandsma Jan 16 '15 at 18:26
  • @HennoBrandsma: Yes the iteration is done differently but after all both give rise to some sort of dense chain of neighborhoods, or? – C-star-W-star Jan 16 '15 at 19:25
  • @BrianM.Scott: But isn't the constructed bump actually the same comparing this one similar to the one on proofwiki with the lower approximation by Rudin? Rudin just gives an intermediate step by semi continuity, or? – C-star-W-star Jan 16 '15 at 19:27
  • @Freeze_S: Might be, might not be: it depends on the choice of sets $U(r)$ and $V_s$. For that matter, two instantiations of either of the constructions could yield different functions, though all would have the desired properties. But that’s beside the point: the point is that the proofs are very different in style. And I much prefer the more topologically written version. – Brian M. Scott Jan 16 '15 at 19:28
  • After all both are nice! :) (+1) – C-star-W-star Jan 16 '15 at 19:29
  • @BrianM.Scott: Ah I guess you mean the way of how to prove continuity, ok. – C-star-W-star Jan 16 '15 at 19:33
  • @Freeze_S: Yes, though I also think that using just the dyadic rationals makes the construction of the open sets just a little simpler. – Brian M. Scott Jan 16 '15 at 19:34
  • @HennoBrandsma: The iteration in both cases is done iteratively so constructive and work for all subset of the rationals not only sufficient ones as the dyadics: The main point is to consider a countable set! The construction then works in all cases exactly the same: $q_{i_-}\leq q_{n+1}\leq q_{i_+}:\overline{V_{i_-}}\subseteq V_{n+1}\subseteq\overline{V_{n+1}}\subseteq V_{i_+}$ More precisely the choice is done explicitely by: $q_{i_-}:=\max{q_i\leq q_{n+1}:1\leq i\leq n},q_{i_+}:=\min{q_{n+1}\leq q_i:1\leq i\leq n}$ – C-star-W-star Jan 18 '15 at 11:01
  • @BrianM.Scott: I see now why you not only think it is more direct but even more topological: In all three cases one has instantenously the open sets from the previous iteration. I'm sorry for the disruption. =( – C-star-W-star Jan 18 '15 at 11:15
  • @BrianM.Scott: Would you mind if I outsource the construction of the chain of open sets into the question? (This way your answer focuses on the topological proof of continuity.) – C-star-W-star Feb 16 '15 at 13:54
  • I was confused with the continuity part of this proof. seems the definition used here is like Any point $x_0\in X$, and any open interval $(c, d)$ containing the point $f(x_0)$, then there is a neighborhood $U$ of $x_0$ with $f(U)\subseteq (c,d)$. But doesn't we have to show the preimage of open is open? @BrianM.Scott – WhyMeasureTheory Dec 22 '21 at 10:03
  • @WhyMeasureTheory: It’s a basic fact (and easy to prove) that the boldface condition is equivalent to the statement that $f^{-1}[(c,d)]$ is open. – Brian M. Scott Dec 22 '21 at 19:37
  • From $f^{-1}[(c, d)]$ to $f(U) \subseteq(c, d)$ is easy becasuse open set contain ngbd. But from $f(U) \subseteq(c, d)$ how I argue that $f^{-1}[(c, d)]$ is open? @BrianM.Scott Here I was stuck. Let $X={0,1,2}$ with topology $\tau_{X}={\varnothing,{0},{1,2}, X}$, and $Y={a, b, c}$ with the discrete topology. The map $0 \mapsto a, 1 \mapsto b, 2 \mapsto c$ is continuous at 0 , but the preimage of the open set ${a, b}$ containing $a$ is ${0,1}$, which is not open. It seems only ngbd definition is working but I really confused with open and $f(U)\subseteq (c,d)$ one. – WhyMeasureTheory Dec 23 '21 at 08:52
  • 1
    @WhyMeasureTheory: Let $X$ be a topological space and $f:X\to\Bbb R$; then $f$ is continuous iff for each $x\in X$ and open interval $(c,d)$ around $f(x)$ there is an open nbhd $U$ of $x$ such that $f[U]\subseteq(c,d)$. For the non-trivial direction suppose that the condition is satisfied, and let $(c,d)$ be any open interval in $\Bbb R$. Let $W=f^{-1}[(c,d)]$; then each $x\in W$ has an open nbhd $U_x$ such that $f[U_x]\subseteq(c,d)$. Clearly for each $x\in W$ we have $x\in U_x\subseteq W$, so $W=\bigcup_{x\in W}U_x$, and $W=f^{-1}[(c,d)]$ is therefore open in $X$. – Brian M. Scott Dec 23 '21 at 20:34
1

Disclaimer: This is community wiki; feel free to correct, edit, expand etc.! ;)

(This is a slight modification taken from Rudin, Real and Complex Analysis.)

Regard for simplicity: $$C:=B\subseteq A^\complement=:U$$

Denote for construction the rationals: $$S:=\mathbb{Q}\cap[0,1]$$

By normality one can find inductively open $V_s$ for $s \in S$ such that $$C\subseteq V_1, \quad \overline{V_0}\subseteq U,\quad V_s\subseteq\overline{V_{s'}} \text{ for } s>s' .$$

Consider the upper and lower semicontinuous functions: $$\underline{h}_s:=s\chi_{V_s}:\quad\underline{h}_s^{-1}(a,\infty)=\varnothing,V_s,\Omega\in\mathcal{T}$$ $$\overline{h}_s:=(1-s)\chi_{\overline{V_s}}+s:\quad\overline{h}_s^{-1}(-\infty,b)=\varnothing,\overline{V_s}^\complement,\Omega\in\mathcal{T}$$

Pointwise suprema resp. infima preserve lower resp. upper semicontinuity: $$\underline{h}:=\sup_{s\in S}\underline{h}_s:\quad\underline{h}^{-1}(a,\infty)=\{\omega:\underline{h}(\omega)>a\}=\bigcup_{s\in S}\{\omega:\underline{h}_s(\omega)>a\}=\bigcup_{s\in S}\underline{h}_s^{-1}(a,\infty)\in\mathcal{T}$$ $$\overline{h}:=\inf_{s\in S}\overline{h}_s:\quad\overline{h}^{-1}(-\infty,b)=\{\omega:\overline{h}(\omega)<b\}=\bigcup_{s\in S}\{\omega:\overline{h}_s(\omega)<b\}=\bigcup_{s\in S}\overline{h}_s^{-1}(-\infty,b)\in\mathcal{T}$$

They approach each other as by contradiction: $$\underline{h}(\omega)<s<s'<\overline{h}(\omega)\implies\omega\in\overline{V_{s'}}\subseteq V_s\not\owns\omega$$ $$\underline{h}_s(\omega)>\overline{h}_{s'}(\omega)\implies\omega\in V_s\subseteq\overline{V_s}\subseteq V_{s'}\subseteq\overline{V_{s'}}\not\owns\omega\quad(s>s')$$

So they together become continuous: $$h:=\overline{h}=\underline{h}:\quad h^{-1}(c-\varepsilon,c+\varepsilon)=h^{-1}(-\infty,c+\varepsilon)\cup h^{-1}(c-\varepsilon,\infty)\in\mathcal{T}$$

Thus the limiting function is the desired bump: $$h(C)\subseteq h(V_1)\equiv1,\,h(U^\complement)\subseteq h(\overline{V_0}^\complement)\equiv0$$

Paul Frost
  • 76,394
  • 12
  • 43
  • 125
C-star-W-star
  • 16,275
1

(Various versions of Urysohn’s lemma)
Dugundji = Dugundji, J.: Topology, Boston: Allyn and Bacon, 1966.
Pervin = Pervin, W.J.: Foundations of General Topology, New York, Academic Press, 1964.
Pontryagin = Pontryagin, L.: Topological Groups, translated from the Russian by Emma Lehmer, Princeton: Princeton University Press, 1946.
Rudin = Rudin, W.: Real and Complex Analysis, 2nd ed., New York: McGraw-Hill, 1974.

A. Pontryagin [p.46, l.$-$16]: $R$ is a compact regular topological space satisfying the second axiom of countability; the key ideas: Pontryagin [p.46, l.$-$11--l.$-$8].
B. Dugundji [p.146, Theorem 4.1 (1) $\Rightarrow$ (2)]: $Y$ is normal.
C. Pervin [p.158, Urysohn's lemma]: $X$ is normal.
D. Rudin [p.40, l.14]: $X$ is a locally compact Hausdorff space.

Remark 1. The proof of A is the primitive model. Regular Lindel$\ddot{\text{o}}$f $\Rightarrow$ Normal [Dugundji [p.311, diagram]], so $B\Rightarrow A$. Thus, B is stronger than A. A stronger hypothesis implies stronger result, so $\bar{G}_r\cap F=\varnothing$ [Pontryagin [p.47, l.8]] is stronger than $U(r)\cap B=\varnothing$ [Dugundji [p.147, l.13]]. The proof method given in A and B is the same.
Remark 2. We may generalize the construction given in Pervin [p.158, l.$-$12--l.$-$2] by replacing $\{r_n\in \mathbb{N}|n\ge 3\}$ with any countable dense set $D$ of real numbers in $(0,1)$. Then the construction given in Dugundji [p.147, l.10--l.26] will be a special case of this generalized construction.
Remark 3. Locally compact $\Rightarrow$ Completely regular [Dugundji [p.311, diagram]], so we cannot use B or C to prove D. A remedy is to require that one of the two closed sets be compact. Rudin [p.38, Theorem 2.7] corresponds Dugundji [p.144, 3.2(2)]. The construction given in Rudin [p.40, l.$-$9--p.41, l.4] is similar to that given in Pervin [p.158, l.$-$12--l.$-$2] except that $r<s \Rightarrow c(G_r)\subset G_s$ [The larger a rational number is, the larger the corresponding set is; Pervin [p.158, l.$-$11]] is different from Rudin [p.41, (3)][The larger a rational number is, the smaller the corresponding set is]. The continuity of $\mathfrak{g}$ [Pervin [p.158, l.$-$1; p.159, l.9]] is proved by a direct argument, while the continuity of $f$ [Rudin [p.41, l.5--l.12]] is proved by an indirect [through semiconcontinuity] and more generalized argument; the direction of generalization: continuity $\rightarrow$ semicontinuity.

For the applications of Urysohn’s lemma, see Example 6.169 and Example 6.170 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/mathematical-methods.

Li-Chung Wang
  • 116
  • 3