# Thread: Why is this set open?

1. ## 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?

$T_es^t$

2. 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?!

3. Originally Posted by CSM
Does the Latex work?!
If you use $$...$$ tags.
$$\left\lfloor x \right\rfloor \;\& \,\left\lceil x \right\rceil$$ gives $\left\lfloor x \right\rfloor \;\& \,\left\lceil x \right\rceil$

4. 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?

5. Like D28, I cannot read the OP either.
I tried to edit the LaTeX, but you changed my edit.
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.
8. $U_x$ 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).