Why is this set open?

• May 5th 2011, 06:35 AM
CSM
Why is this set open?
Let X be a completely regular space. And subsets K ⊂ X compact and C ⊂ X closed, with K ∩ C = ∅.
(i) Show that there are functions fx ∈ C(X, [0, 1]) with fx(x) = 1 and fx = 0 on C, where x ∈ K.
(ii)Use the functions fx to make an open cover of K
(iii)Use compactness to make a function g ∈ C(X, [0,∞)) which is g = 0 on C and g > 1/2 on K.
(iv) Change g to get a f ∈ C(X, [0, 1]) with f = 1 on K and f = 0 on C.
NOTE: X doesn't have to be normal (T4)!

X is completely regular and C ⊂ X is closed so for every x not in C there is a fx
C(X, [0, 1]) with f(x) = 1 and f = 0 on C. We do this for every x ∈ K and deﬁne $U_x =f_x^{-1}((\frac{1}{2}, 1])\subset X$. Ux is open and contains x. So {Ux | x ∈ K} is an open cover of K. K is compact, so there are x1, . . . , xn ∈ K such that Ux1 ∪ · · · ∪ Uxn ⊃ K. We define g(x) = fx1(x) + · · · + fxn(x) or g(x) = max{fx1(x), . . . , fxn(x)}. For both definitions it follows that g ∈ C(X, [0,∞)) and g = 0 on C. Suppose x ∈ K. Then x ∈ Ui for some i, so fi(x) > 1/2.
So g(x) > 1/2. Let h ∈ C([0,∞), [0, 1]) be such that h(0) = 0 and h(t) = 1 for t ≥ 1/2. Then f = h ◦ g has the desired properties.

Question: Why is Ux open? What are the other possibilities of making an open cover of K by means of using the fx? Is this the most natural choice?

$T_es^t$
• May 5th 2011, 06:58 AM
CSM
Quote:

Originally Posted by CSM
Let X be a completely regular space. And subsets K ⊂ X compact and C ⊂ X closed, with K ∩ C = ∅.
(i) Show that there are functions fx ∈ C(X, [0, 1]) with fx(x) = 1 and fx = 0 on C, where x ∈ K.
(ii)Use the functions fx to make an open cover of K
(iii)Use compactness to make a function g ∈ C(X, [0,∞)) which is g = 0 on C and g > 1/2 on K.
(iv) Change g to get a f ∈ C(X, [0, 1]) with f = 1 on K and f = 0 on C.
NOTE: X doesn't have to be normal (T4)!

X is completely regular and C ⊂ X is closed so for every x not in C there is a fx
C(X, [0, 1]) with f(x) = 1 and f = 0 on C. We do this for every x ∈ K and deﬁne $U_x =f_x^{−1}((\frac{1}{2}, 1])\subset X$. Ux is open and contains x. So {Ux | x ∈ K} is an open cover of K. K is compact, so there are x1, . . . , xn ∈ K such that Ux1 ∪ · · · ∪ Uxn ⊃ K. We define g(x) = fx1(x) + · · · + fxn(x) or g(x) = max{fx1(x), . . . , fxn(x)}. For both definitions it follows that g ∈ C(X, [0,∞)) and g = 0 on C. Suppose x ∈ K. Then x ∈ Ui for some i, so fi(x) > 1/2.
So g(x) > 1/2. Let h ∈ C([0,∞), [0, 1]) be such that h(0) = 0 and h(t) = 1 for t ≥ 1/2. Then f = h ◦ g has the desired properties.

Question: Why is Ux open? What are the other possibilities of making an open cover of K by means of using the fx? Is this the most natural choice?

$T_es^t$

Does the Latex work?!
• May 5th 2011, 07:11 AM
Plato
Quote:

Originally Posted by CSM
Does the Latex work?!

If you use $$...$$ tags.
$$\left\lfloor x \right\rfloor \;\& \,\left\lceil x \right\rceil$$ gives $\left\lfloor x \right\rfloor \;\& \,\left\lceil x \right\rceil$
• May 5th 2011, 11:27 AM
Drexel28
Quote:

Originally Posted by CSM
Let X be a completely regular space. And subsets K ⊂ X compact and C ⊂ X closed, with K ∩ C = ∅.
(i) Show that there are functions fx ∈ C(X, [0, 1]) with fx(x) = 1 and fx = 0 on C, where x ∈ K.
(ii)Use the functions fx to make an open cover of K
(iii)Use compactness to make a function g ∈ C(X, [0,∞)) which is g = 0 on C and g > 1/2 on K.
(iv) Change g to get a f ∈ C(X, [0, 1]) with f = 1 on K and f = 0 on C.
NOTE: X doesn't have to be normal (T4)!

X is completely regular and C ⊂ X is closed so for every x not in C there is a fx
C(X, [0, 1]) with f(x) = 1 and f = 0 on C. We do this for every x ∈ K and deﬁne $U_x =f_x^{−1}((\frac{1}{2}, 1])\subset X$. Ux is open and contains x. So {Ux | x ∈ K} is an open cover of K. K is compact, so there are x1, . . . , xn ∈ K such that Ux1 ∪ · · · ∪ Uxn ⊃ K. We define g(x) = fx1(x) + · · · + fxn(x) or g(x) = max{fx1(x), . . . , fxn(x)}. For both definitions it follows that g ∈ C(X, [0,∞)) and g = 0 on C. Suppose x ∈ K. Then x ∈ Ui for some i, so fi(x) > 1/2.
So g(x) > 1/2. Let h ∈ C([0,∞), [0, 1]) be such that h(0) = 0 and h(t) = 1 for t ≥ 1/2. Then f = h ◦ g has the desired properties.

Question: Why is Ux open? What are the other possibilities of making an open cover of K by means of using the fx? Is this the most natural choice?

$T_es^t$

Friend, I tried to figure out how you defined $U_x$ but the code is indecipherable. Could you possibly use the tex commands?
• May 5th 2011, 12:25 PM
Plato
Like D28, I cannot read the OP either.
I tried to edit the LaTeX, but you changed my edit.
Maybe you can find that webpage useful.
• May 5th 2011, 01:47 PM
Tinyboss
So Urysohn's Lemma says that in a normal space, every pair of disjoint closed sets can be separated with a function. This problem wants you to show that, if we only require the space to be completely regular, rather than normal, we can still get the result when one set is closed and the other is compact. (An example of the metatheorem that compact sets are like "fat points"--if we can separate a point from a closed set, then we can separate a compact set from a closed set.)

As you said, we get the $f_x$'s straight from the definition of completely normal. I suspect your definition is $U_x=\{f_x^{-1}(1/2,1]\}$, which is indeed an open set containing x and missing C. Pick a finite set $\{x_1,\dots,x_k\}$ of points in K such that $\bigcup_{i=1}^kU_{x_i}\supset K$ (by compactness). Then just define $F(x)=\sum_{i=1}^kf_{x_i}}(x)$, which is finite, zero on C, and at least 1/2 on K. The last transformation is trivial.
• May 6th 2011, 03:31 AM
CSM
Okay thanks for your replies, the latex is fixed now. Your definition of the set is correct. My question now is, why is U_x open and why choose that construction of U_x, is it the most straightforward choice? Is it trivial? Enlighten me please.
• May 6th 2011, 04:28 AM
Tinyboss
$U_x$ is open because it's the inverse image of the open set $(1/2,1]$ (remember, it's a subset of the range [0,1], so it is indeed open) under $f_x$, which is continuous (we're guaranteed that by complete regularity of the space).
• May 7th 2011, 07:25 AM
CSM
Thanks. Looking back it is pretty easy, thanks to your input. I took the inverse of [1] which obviously isn't open. So everything went wrong from there on.