Posted by: Dave Richeson | October 2, 2008

## Kuratowski’s closure-complement theorem (solution)

Stop!  This post contains spoilers.  This page has the solution to the problem posed in yesterday’s post.

We challenged you to find a set $A\subset \mathbf{R}$ from which we can make as many new sets as possible using only the closure and complement operations.

In 1922 Kuratowski proved the following theorem.

Theorem. At most 14 sets can be obtained from $A\subset\mathbf{R}$ (including $A$ itself) using the closure and complement operations.

This is often called Kuratowski’s closure-complement theorem or Kuratowski’s 14-set theorem. (Note: he proved this for general topological spaces, not just $\mathbf{R}$.)

In fact, we can achieve this value.  An example of such a set is…

$A=(-\infty,0)\cup (0,1)\cup(\mathbf{Q}\cap(1,2))\cup\{3\}$

We leave it to the reader to verify that we can produce 14 sets from $A$ using closures and complements.

James Fife gave a nice, readable proof (subscription required) of this theorem in 1991 (Mathematics Magazine, Vol. 64, No. 3 (Jun., 1991), pp. 180-182).

## Responses

1. An interactive solution to the Kuratowski Closure-Complement Theorem can be found at

http://www.kuratowski.com

2. That’s a great applet. Thank you for sharing it.

3. […] Kuratowski’s closure-complement theorem (solution). […]