Frutiger has similarities to Univers. But where Univers is an engineered sans like Helvetica,
Frutiger has more humanistic roots. If I remember correctly, Adrian Frutiger designed his
eponymous typeface on a commission to redo the signage for the Charles de Gaulle Airport of Paris.
This was maybe 1967 or sometime around then. So it's been around a while, but I don't think I've
seen it in any publications from earlier than the 1990s. I don't know why it would have languished
for so long, because it's another masterpiece. Eminently legible and, furthermore, really
pleasant to read. It has a softer effect than Univers. I find that it makes almost anything
fun to read. Nowadays the Swiss government uses it for a lot of official signage
such as the Postbus logo. This is another one which, morally speaking,
should be way more popular than Arial.
THE POWER SET OF A SET IS REALLY BIG AND THIS IS WHY
Note: I've come up with a much more streamlined (and precise) way to demonstrate the
following result, namely that the power set of a countably infinite set is uncountable.
I'll also add the related result that the collection of finite subsets of a
countable set is countable. All this coming soon.
Why is the power set of a set bigger than the original set? What is a power set, you ask?
The power set P(A) of a set A is the set of all the subsets of A, including
the empty set and A itself. For finite sets the answer to the first question is
obvious, both intuitively and
if you know how to compute the size of a power set: if set A has n elements, then P(A)
has 2^n elements. For instance, the set {1,2,3} has power set {Ø,
{1}, {2}, {3}, {1,2}, {1,3}, {2,3},
{1,2,3}}. This has 8 elements because each element of {1,2,3} can be either
in or out of one of its subsets, giving a total of 2·2·2 = 2^3 = 8 subsets.
If you're looking at an infinite set U, then clearly P(U) is going to be infinite as well.
But how do we know it's bigger? How can an infinite set ever be "bigger" than another infinite set?
Well, the way two infinite sets are determined to be of the same cardinality ("size") is to find
a one-to-one correspondence between each of their elements. Take the set of natural numbers N =
{1, 2, 3, 4, 5, ...} as an example. Define -N as {-n : n is in N} =
{-1, -2, -3, -4, -5, ...}. Intuitively, N and -N have the same cardinality, and indeed
it is trivial to find a bijection (one-to-one correspondence) between them; this is it: define
f from N to -N by f(n) = -n for all n in N.
Each element of -N is mapped to by f from exactly one element of N, and the range of
f (the set of objects mapped to by f) is all of -N.
There is no bijection between N and P(N). These two sets do not have the same cardinality.
For the longest time though, I thought there was such a bijection g, defined as follows.
For any n in N, consider the representation of n in binary. For instance, 1 stays 1 but 2 becomes 10,
3 becomes 11, 4 becomes 100, and so on. Now associate this representation
with the set of natural numbers determined from it this way: if in the kth place in the notation
there is a 1, include k in the set; if there is a 0, exclude k. So for example, the set associated
to 50 = 110010 (in binary) would be {2, 5, 6}.
g(n), then, for all n in N, is the set associated in this way from the binary representation of n - 1.
We see that g(1) = Ø, g(2) = {1}, g(3) = {2}, g(4) = {1, 2},
and so on. No two natural numbers map to the same set. Another way to say this is that
for all m and n in N, if g(m) = g(n), then m = n. This result shows that g is injective, one of the
two conditions for bijection. The other condition is that g be surjective, that is, that for all
elements A of the target set - in this case, all sets A that are subsets of N and hence elements
of P(N) - there is an element a of the original set - N - such that g(a) = A. That is, each element
of P(N) is mapped to by g from some element of N.
g is not surjective.
Maybe you can already see why this is. If not, that's totally cool, because there elapsed oh maybe
five years or more between the time I first constructed g and the time I figured out why it's not
surjective, over this past spring break. So here's the reason. I find it pretty obvious now.
If we look at a finite subset A of N, then there is indeed a natural number a that g maps to it.
The prob is, there are a lot
of infinite subsets of N as well (such as N itself, as well as {2, 3, 4, 5, ...},
{1, 3, 4, 5, ...}, and {3, 4, 5, ...} - there are tons of 'em).
Do you see where this is going? Well, what natural number maps to N? or to any of the other
infinite subsets I just listed? The answer: There is none. Only infinity would map to any of these,
and infinity is not an element of N. Even we considered it to be an element of N, if we chose
g(infinity) = U for some specific infinite subset U of N, then all the other infinite subsets would
have nothing that we could map to them. So clearly, g is not surjective - it does not cover all the
elements of P(N); in particular it does not cover any infinite subset of N. It does, however, cover
all finite subsets of N, and so is a bijection between N and the set F(N) of finite subsets of N. So
these two sets have the same cardinality. But P(N) has a greater cardinality than either, because
of the gigantic number of infinite subsets of N. There you have it. Does this make sense? Let me
know if you want something in there clarified. I hope you find this pretty cool like I do.
. . . Addendum: Let me add something I forgot to mention, it's pretty important. So
important that the explanation up till now is incorrect without it. You might
have considered that the set of infinite subsets of N, which is P(N)\F(N) - this notation denotes
the set of all elements of P(N) that are not in F(N) - is in fact in a one-to-one
correspondence h with the set of finite subsets N, in this way: for every finite subset A of N,
let h(A) = N\A. This is a valid bijection. So the collection of all such infinite subsets
N\A, call it I(N), has the same cardinality as F(N), and so if it is indeed equal to P(N)\F(N)
as we think it may be, then this is a problem. And this is why:
if two sets both have the same cardinality as N, then their union (the set of all
elements of both sets) also has the same cardinality as N. (Try to figure this out by finding a
bijection between N and the union of N and -N, i.e. the set of nonzero integers.) So if
I(N) = P(N)\F(N), then the union of F(N) and P(N)\F(N),
which would be P(N), has the same cardinality as N.
Luckily for the purposes of this proof not falling apart,
I(N) is not the same set as P(N)\F(N); P(N)\F(N) has a
whole lot of elements that I(N) does not. After all, I(N) is the set of all infinite subsets of N
with a finite number of elements missing. What about the infinite subsets of N with an infinite
number of elements missing? There are plenty of these; to name a few, there is the set of even
natural numbers, the set of odd natural numbers, and the set of multiples of 3. In fact, such
sets are so legion in P(N) that they completely overwhelm F(N) and I(N), and there is no bijection
from N that can handle them. This is the full-disclosure reason why the cardinality of P(N) is
greater than that of N. Thank you and goodnight.
This part's not started yet. One thing I want to discuss here is why sound color is the most
important factor in how much I like a song.
