Assume that {xi} converges to x. Let N be a neighborhood of πβ(x) for some β in the index set of ∏ Xα. Then U ⊂∏ Xα where πβ(U) = N and πα(U) = Xα for α≠β is a neighborhood of x since it is open in the product topology and contains x. Hence there are infinitely many xj of {xi} which are also in U, however each xj has that πβ(xj) ∈ πβ(U) = N. Thus N contains infinitely many points of {πβ(xi)}i, so πβ(xi) → πβ(x).
Conversely, assume that for all β in the index set of ∏ Xα the sequence {πβ(xi)} converges to πβ(x). Let V be a neighborhood of x. Then there is a basis element B of ∏ Xα with x ∈ B ⊂ V . Being in the product topology, only finitely many coordinates of B are sets which are not Xα. Letting πα1(B),…,παm(B) be these sets, we have, by assumption, that there exist N1,…,Nm such that all n ≥ Nj has that παj(xn) ∈ παj(B). Thus if we set
|
| (P-1.1) |
then πα(xn) ∈ πα(B) for all α in {α1,…,αm} and n ≥ N. From this we get that xn ∈ B ⊂ V for all n ≥ N, which implies the convergence of {xi} to x.
Does the same hold for the box topology? The latter half of the above proof is dependent on the finititude of the number of coordinate sets in a given basis element of the product topology on ∏ Xα. As exemplified by the following scenario, the box topology may disallow such an N as in P-1.1 above.
Let ∏
Xα be ℝω with the box topology and define the sequence {xi} by xi = (1 ∕ i,2 ∕ i,3 ∕ i…). So for any coordinate j,
(πj(x1),πj(x2),πj(x3),…) = (j ∕ 1,j ∕ 2,j ∕ 3,…) and certainly the right hand side converges to 0. Thus x = (0,0,0,…) will have
the property that for each j, πj(xi) → πj(x). However, for the neighborhood B = (-1,1) × (-1,1) × (-1,1) ×
of
x = (0,0,0,…) there exists no N such that all n ≥ N has xn ∈ B since πN(xN) = N∕ N = 1 ⁄∈ (-1,1) and thus
xN ⁄∈ B.
prod ⊂ uniform ⊂ box. So it only remains to be shown that the box topology has a basis element for which no basis
element of the uniform topology is contained within, and that the uniform topology contains a basis element for which no
basis element of the product topology is contained within.
In the case of ℝJ where J is infinite, set B to be a basis element of the box topology defined as follows: for each n ∈ ℤ+
there exists a unique α ∈ J such that πα(B) = (-1 ∕ n,1 ∕ n) and if α′ is not such an α, then πα′(B) = ℝ. Note that for the
following example the α ∈ J for which πα(B) = ℝ and for which πα(B)≠ℝ do not matter, just that there are a countably
infinite number of them. So for 0 ∈ B, the sequence of all zeros, there is no open ball Bρ(0,r) of the uniform
topology such that 0 ∈ Bρ(0,r) ∈ B since for any n such that 1 ∕ n < r, an α ∈ J can be found such that
Bd(0,r) ⁄⊂ πα(B) = ( - 1 ∕ n,1 ∕ n), since, say, 
is in Bd(0,r) but not in ( - 1 ∕ n,1 ∕ n). Thus the uniform and box topologies
are different.
To see that the uniform topology on ℝJ is different from the product topology, we need only look at an open ball of the uniform topology with radius r < 1. This will be the set ∏ αBd(xα,r) for some center (xα), but no basis element of the product topology can be contained within this set since no basis element B of the product topology can have more than a finitie number of α ∈ J with πα(B)≠ℝ.
With this definition we see that d′ is a metric by
+d(xn,yn)+d(y1,z1)+
+d(yn,zn) =
d(x1,y1) + d(y1,z1)
+ 
+ 
d(xn,yn) + d(yn,zn)
≥ d(x1,z1) + 
+ d(xn,zn) = d′(x,z), where d is the
euclidean metric on ℝ.Now that we have that d′ is indeed a metric we can compare its induced topology with that of the square metric, ρ, on ℝn via Munkres’ Lemma 20.2. Clearly for any x,y ∈ ℝn
 | (P-3.2) |
and
 | (P-3.3) |
By Equation P-3.2, inside of any ϵ-ball Bρ(x,ϵ) is one Bd′(x,ϵ). So Munkres’ Lemma 20.2 tells us that the topology induced by d′ is finer than the one induced by ρ.
By Equation P-3.3, any Bd′(x,ϵ) has a Bρ(x,ϵ ∕ n) contained within it. Again turning to Munkres’ Lemma 20.2, we get that the topology induced by ρ is finer than that of d′.
Given the above two results, d′ and ρ induce the same topology on ℝn, that is, the usual topology.
Sketching. Figure 1 shows a ball for the metric d′ in ℝ2. It is no particular ball.
be a collection of connected subspaces of X such that An ∩An+1≠∅. For
our base case, A1 is connected by assumption. Assume that for 1 through n - 1 we have ⋃
i=1,…,n-1Ai is
connected. Assume for later contradiction that C,D separate ⋃
An. By Munkres Lemma 23.2, ⋃
i=1,…,n-1Ai is
contained within C or D, so without loss of generality allow ⋃
i=1,…,n-1Ai ⊂ C. By the same Lemma, An
must also be contained within one of C or D, however, An has nonempty intersection An-1 and therefore
with ⋃
i=1,…,n-1Ai as well. Thus since ⋃
i=1,…,n-1Ai is contained in C, An must also be, thereby yielding
that D actually is the emptyset. This contradicts C and D separating ⋃
An and therefore ⋃
An must be
connected.
Let C be the disc centered at the origin, {(x,y) ∈ ℝ2 : |x| + |y|≤1 ∕ 2}. Define d (our metric in question) on ℝ2 by the following
This 2 acts as our infinity, since the standard bounded metric has a maximum value of 1; regarding our original motivation, this metric makes the border of C “impenetrable” resulting in “infinite” distance of 2.
Metric Proof. Despite calling it a metric already, we have yet to prove that d is indeed one. Alas it is:
The Finale.
Now with this metric d we can set B1 = Bd
(0,0),3 ∕ 4
and B2 = Bd
(1 ∕ 2,0),7 ∕ 8
. So both (0,0) and (1 ∕ 2,0) are in C,
and clearly B1 = C and B2 ⊂ C since they cannot contain any points outside of C due to their radii being less than 2.
However despite the fact that B2 has a larger radius of 7 ∕ 8, it doesn’t contain the point (-1 ∕ 2,0) of C since this is of
distance 1 away from the center of B2, (0,1 ∕ 2). Hence B2 ⊊ B1.
