# Can Category Theory Serve as the Foundation of Mathematics?

Usually the foundation of mathematics is thought of as having two pillars: mathematical logic and set theory.  We briefly discussed mathematical logic and the foundation of mathematics in the story of Gödel, Brouwer, and Hilbert. The story of set theory is one of the most exciting in the history of mathematics, and its main hero was Georg Cantor, who discovered that there are many types of “infinity.”

Mathematical logic was always considered a very abstract part of mathematical activity, related to philosophy and quite separate from applications of mathematics. With the advent of computers, however, this perception completely changed. Logic was the first, and for many years, the main mathematical discipline used in the development of computers, and to this day large parts of computer science can be regarded as “applied logic.”

While mathematical logic and set theory indeed make up the language spanning all fields of mathematics, mathematicians rarely speak it. To borrow notions from computers, basic mathematical logic can be regarded as the “machine language” for mathematicians who usually use much higher languages and who do not worry about “compilation.” (Compilation is the process of translating a high programming language into machine language.)

The story of Category Theory is markedly different from that of mathematical logic and set theory. It was invented Continue reading

# Gödel, Hilbert and Brouwer

Is mathematics a consistent theory? Or, rather, is there a danger of finding a correct mathematical proof for a false statement like “0 = 1″?  These questions became quite relevant at the end of the nineteenth century, when some mathematical truths dating back many centuries were shattered and mathematicians started to feel the need for completely rigorous and solid foundations for their discipline.  Continue reading

# A Diameter problem (7): The Best Known Bound

### Our Diameter problem for families of sets

Consider a family $\cal F$ of subsets of size d of the set N={1,2,…,n}.

Associate to $\cal F$ a graph $G({\cal F})$ as follows: The vertices of  $G({\cal F})$ are simply the sets in $\cal F$. Two vertices $S$ and $T$ are adjacent if $|S \cap T|=d-1$.

For a subset $A \subset N$ let ${\cal F}[A]$ denote the subfamily of all subsets of $\cal F$ which contain $A$

MAIN ASSUMPTION: Suppose that for every $A$ for which ${\cal F}[A]$ is not empty $G({\cal F}[A])$ is connected.

We will call a family satisfying this assumption “hereditarily connected”.

MAIN QUESTION:   How large can the diameter of $G({\cal F})$ be in terms of $d$ and $n$?

We denote the answer by $F(d,n)$.

For $v \in \{1,2,\dots,n\}$ let ${\cal F'}[v]$ be the family obtained from ${\cal F}[\{v\}]$ by removing $v$  from every set. Since $G({\cal F}[v]) = G({\cal F}' [v])$, the diameter of  $G({\cal F}[\{v\}])$ is at most $F(d-1,n-1)$.

### 8. A slight generalization

Let ${\cal F}$ be an hereditarily connected family of $d$-subsets of a set $X$. Let $Y$ be a subset of $X$. The length of a path of sets $S_1,S_2,\dots, S_t$ modulo Y  (where $|S_i \cap S_{i+1}|=d-1$ for every $i$) is the number of $j, 1 \le j such that both $S_j$ and $S_{j+1}$ are subsets of $Y$. (In other words, in $G({\cal F})$ we consider edges between subsets of Y as having length 1 and other edges as having length 0.)

Let $T(d,n)$ be the largest diameter of an hereditarily connected family of $d$-subsets of an arbitrary set $X$ modulo a set Y , $Y \subset X$ with $|Y|=n$.

Since we can always take $X=Y$ we have $F(d,n) \le T(d,n)$.

### 9.  A quasi-polynomial upper bound

We will now describe an argument giving a quasi-polynomial upper bound for $T(d,n)$. This is an abstract version of a geometric argument of Kleitmen and me.

Let ${\cal F}$ be a hereditarily connected family of $d$-subsets of some set $X$, let $Y \subset X$, $|Y|=n$, and let $S$ and $T$ be two sets in the family.

Claim: We can always either

1) find paths of length at most $T(d,k)$  modulo from $S$ to $d$-subsets of $Y$ whose union has more than $k$ elements.

or

2) we can find a path of this length $T(d,k)$  modulo Y   from $S$ to $T$.

Proof of the claimLet $Z$ be the set of elements from $Y$ that we can reach in $T(d,k)$ steps modulo Y   from $S$. (Let me explain it better: $Z$ is the elements of $Y$ in the union of all sets that can be reached in $T(d,k)$ steps modulo Y from $S$. Or even better: $Z$ is the intersection of $Y$ with the union of all sets in $\cal F$ which can be reached from $S$ in $T(d,k)$ steps modulo Y. )

The distance of $S$ from $T$ modulo Z  is at most $T(d,|Z|)$.

Now, if $|Z|>k$ we are in case 1).

If $|Z| \le k$ then there is a path from $S$ to $T$ modulo Z of length $T(d,k)$. If this path reaches no set containing a point in $Y \backslash Z$ we are in case 1).  (Because this path is actually a path of length $T(d,k)$ from $S$ to $T$ modulo Y).  Otherwise, we reached via a path of length $T(d,k)$ modulo Y from $S$ a set containing a point in $Y \backslash Z$, in contradiction to the definition of $Z$.  Walla.

Corollary: $T(d,n) \le 2T(d,n/2)+T(d-1,n-1)$.

By a path of length $T(d,n/2)$ modulo Y  we reach from $S$ at least $n/2$ elements in $Y$, (or $T$).  By a path of length $T(d,n/2)$ modulo Y  we reach from $T$ at least $n/2$ elements in $Y$, (or $S$). So unless we can go from $S$ to $T$ in $T(d,n/2)$ steps modulo Y  we can reach more than $n/2$ elements from both $S$ and $T$ by paths of length $T(d,n/2)$ modulo Y ,hence there is some element we can reach from both.

In other words in $T(d,n/2)$ steps modulo Y  we go from $S$ to $S'$ and from $T$ to $T'$ so that $S'$ and $T'$ share an element $u$.

But the distance from $S'$ to $T'$ modulo Y  (which is the same as the distance modulo Y  from $S' \backslash u$ to $T' \backslash u$ in ${\cal F}'[\{u\}]$ is at most $T(d-1,n-1)$.  (We use here the fact that $u \in Y$) Ahla!

To solve the recurrence, first for convenience replace $T(d-1,n-1)$ by $T(d-1,n)$. (You get a weaker inequality.) Then write $G(d,n)=T(d,2n)$ to get $G(d,n) \le G(d,n/2)+G(d-1,n)$ and $H(d,x)=G(d,2^x)$ to get $H(d,x) \le H(d-1,x) + H(d,x-1)$ which gives $H(d,x) \le {{d+x} \choose {d}}$ which in turn gives $G(d,n) \le {{log n+d} \choose {d}}$ and $T(d,n) \le n {{log n +d} \choose {d}}$. Sababa!