(i) d1(f, f) = R 1
0
|f(x) − f(x)| dx =
R 1
0
0 dx = 0. And, if f 6= g, then
F(x) = |f(x)−g(x)| is not identically zero. Hence ∃x0 so F(x0) = 2 >
- And by continuity, ∃δ such that ∀x, x0−δ < x < x0+δ, F(x) > . So,
d1(f, g) = R 1
0
F(x) dx =
R x0−δ
0
F(x) dx +
R x0+δ
x0−δ
F(x) dx R 1
x0+δ
F(x) dx ≥
R x0+δ
x0−δ
, dx = 2δ > 0
0 = d1(f, g) = R 1
0
|f(x) − g(x)| dx, implies |f(x) − g(x)| = 0 for all x,
and thus f = g.
(ii) d1(f, g) is symmetric in definition, as |f − g| = |g − f|, and thus
equals d1(g, f).
(iii) d1(f, h)+d1(h, g) = R 1
0
|f(x)−h(x)|+|h(x)−g(x)| dx ≥
R 1
0
|f(x)−
g(x)| dx = d1(f, g). - {fn} : ∀ > 0, ∃N(> 1/) so that ∀n > N, x ∈ R we have 0 <
n
x2+n2 ≤
n
n2 < . Thus, {fn} converges uniformly (and thus also point-wise) to
0 on all of R.
{gn} : For any given x, we can pick a sufficiently large N(>
q
x2
) so
that, n > N implies gn is close to, 1. Thus, {gn} converges point-wise.
But, for any n, if we pick x = n to get gn =
1
2
. This means that {gn}
does not converge uniformly to 1. - Let U = {x : d(x, A) < d(x, B)} and V defined anaglgously, with d
the distance function defined in problem set 2 problem 1. Clearly, U
and V are disjoint and contain A and B respectivly. Now, it remains
to show that they are open. Let x ∈ U, =
1
3
(d(x, B) − d(x, A)), and
y ∈ B(x). Then, d(y, B) − d(y, A) > d(x, B) − d(x, A) − 2 > 0, and
we have y ∈ U and U open. So U and V are as desired. - I claim that for U, V open, U ⊂ U¯ ⊂ V , there exists W open so that
U¯ ⊂ W ⊂ W¯ ⊂ V . To see this, consider U¯ and X − V , disjoint closed
sets. Then there exists W ⊃ U¯, W0 ⊃ X − V open and disjoint. But
then U¯ ⊂ W ⊂ X − W0 ⊂ X −V . Since X − W0
is closed and contains
W, it also contains W¯ . And we have U ⊂ U¯ ⊂ W ⊂ W¯ ⊂ V , as
desired.
1
Now, letting S1 ⊂ X −B, and S0 ⊃ A as produced by X −B and IntA
with the above lemma applied twice. I inductivly create open Sk/2
i one
level of “i” at a time. At each point if q < r, then S¯
q ⊂ Sr. Sk/2
i (k
odd), is generated by the above lemma between S(k−1)/2
i and S(k+1)/2
i .
So of course, all the sets contain the closure of S0 and are contained in
S1.
I define f(x) = inf({1} ∪ {r : x ∈ Sr}). This is set is bounded below
and non empty, so the inf exists. It is clear that f(x) = 0 if x ∈ A,
f(x) = 1 if x ∈ B, and has range [0, 1]. It now remains to show that
f is continuous. I first show f(x) = sup({0} ∪ {r : x ∈ X − S¯
r}). In
doing this, we only need to consider r in the sets that are terminating
binary fractions, as for no other r is Sr defined and thus is it possible
for r to satisfy the condition to be in the sets.
If, r > f(x) then ∃r0, r > r0 > f(x) with x ∈ Sr0 ⊂ Sr ⊂ S¯
r, making
x 6∈ X −S¯
r. As 0 ≤ f(x), we can upperbound the sup with f(x). And,
∀r < f(x) ≤ 1, then ∃r0, r < r0 < f(x) with x 6∈ Sr0 ⊃ S¯
r and thus
x ∈ X − S¯
r. Since all binary fraction 0 < r < f(x) (it is key here that
f(x) is bounded above 1 so that all of these r produce valid Sr) are in
the set, and these are dense in the reals, we have f(x) as a lower bound
for the sup as well. This fails if f(x) = 0, but then, the additional 0
element saves us. Regardless we have now proved the identity.
Next, we show f
−1
([0, r)) is open (r is now any real number). I claim
f
−1
([0, r)) = ∪s<rSs. If x ∈ ∪s<rSs, then ∃s < r such that x ∈ Ss and
thus, f(x) = inf({1} ∪ {a : x ∈ Sa}) ≤ s < r. If x 6∈ ∪s<rSs, then
∀s < r, x 6∈ Ss. Thus, if x ∈ Ss, then s ≥ r; and of course 1 ≥ r.
So, f(x) = inf({1} ∪ {a : x ∈ Sa}) ≥ r. So x ∈ ∪s<rSs if and only
if x ∈ f
−1
([0, r)) making the sets equal; as the union of open sets is
open so is f
−1
([0, r)). The same argument, using the sup definition of f,
shows that f
−1
((r, 1]) is open. As, f
−1
((a, b)) = f
−1
([0, b))∩f
−1
((a, 1]),
this set is open as well. Finally, any open set in the reals is the union
of open balls (one around each point if you like), f
−1 of any open set
is the union of open sets and is thus open. Thus, f is continuous as
desired. - Let X be a topological space and F a collection of closed subsets.
Define G as the complements of sets in F. Then a union of sets in G is
the complement of an intersection of the corresponding sets in F. So
2
F has FIP if and only if G has no finite subcover of X. Also F has the
total intersection property if and only if G does not cover X. Therefore
we are done. - Let S be a sequentially compact subset of a metric space.
(i) Suppose S is not closed. Then ∃x 6∈ S, such that ∀r > 0, Br(x)∩S 6=
∅. Let, an be a point in S ∩ B1/n(x). Clearly, an → x 6∈ S, and thus
all subsequences converge to x 6∈ S and thus do not converge in S,
contradicting the sequential compactness of S.
(ii) Suppose S is not totally bounded. Then let r be a radius such that
there is no r net of S. Now define an as follows: Let a0 be any point in
S and let an ∈ S, such that ∀m < n, d(am, an) ≥ r. Such an an exists
by the lack of an r net of S. But S sequentially compact implies ∃m, n
such that d(am, an) < r which is impossible. So, S must be totally
bounded. - Clearly if E is totally bounded, then it is totally bounded relative to
any metric space containing it. Conversely, if E is totally bounded
relative to X, then there is some finite set of point p1, p2, . . . pn ∈ X,
so that every point in E is within /2 of these points. Without loss
of generality, we may assume that there is some qi ∈ E in each of
these /2 balls, or else we omit the corresponding pi
from the original
enumeration. But now, ∪iB(qi) ⊃ ∪iB/2(pi) ⊃ E, by the triangle
inequality. And thus for every we have an epsilon net of E centered
around points in E, making E totally bounded. - Suppose no such r exists. Then for each r, in particular for each 1/n,
there is some xn such that ∀α, B1/n(xn) 6⊆ Uα. Now, X is compact, so
exists ni ∈ Z
+, x ∈ X, so that xni → x. Now, {Uα} covers X, so ∃α, so
that x ∈ Uα. As these sets are open, ∃r such that Br(x) ⊂ Uα. And,
by convergence, ∃k > 2
r
, so that d(xk, x) <
r
2
. But then, B1/k(xk) ⊂
Br(x) ⊂ Uα, contradicting our construction of {xn}, and implying the
existence of such an r
Discord