Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Okay, so here's the problem:

A function is a decision function if it maps finite length binary strings {0,1}* into codomain {0,1}. Let D be the set of all possible decision functions. Show that D is uncountable.

My attempts so far have been:

Let f be a decision function s.t. $f: \{0,1\}^* \rightarrow \{0,1\}$

Therefore, $f$ is in D, $f(x)$ is in $\{0,1\}$. I also want to say that the input of $f$ must be in the form:

$B_n = \{$ $b_i$ | $b_i \in \{0,1\}\}$

My biggest concern is that this function seems surjective to me, which would indicate that it is countable. But maybe I'm thinking of the wrong part of the problem. I need to show that D is uncountable, so I need to show that the cardinality of D is greater than some other uncountable set. How could I start showing this? What should I be thinking of/about? Any help will be appreciated.

share|cite|improve this question
up vote 3 down vote accepted

HINT: The set of finite binary strings is countably infinite, so there are exactly as many decision functions as there are functions from $\Bbb N$ to $\{0,1\}$.

share|cite|improve this answer
is this where I use Cantor's diagonal argument? – atb Oct 23 '12 at 21:34
@alexthebake: That’s one way to go, yes. Alternatively, if you’ve already proved that the set of functions from $\Bbb N$ to $\{0,1\}$ is uncountable, you can just set up a bijection between these functions and your decision functions. – Brian M. Scott Oct 23 '12 at 21:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.