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This is the question : Prove that the set of all the words in the English language is countble (the set's cardinality is אo) A word is defined as a finite sequence of letters in the English language.

I'm not really sure how to start this. I know that a finite union of countble sets is countble and i think this is the way to start.

Thanks in advance !

Andrés E. Caicedo
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NotSure
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  • This is not the way to start. Look for countability of the set of finite strings. The chosen alphabet is irrelevant as long as it's countable. – AlexR Mar 25 '15 at 19:27
  • A countable union of countable sets is also countable. – Santiago Canez Mar 25 '15 at 19:28
  • @AlexR: the chosen alphabet is irrelevant as long as it's countable. – TonyK Mar 25 '15 at 19:30
  • @TonyK Thanks I noticed that after reading on my found dupe. Edited within the grace period ;) – AlexR Mar 25 '15 at 19:33
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    The set of all English words is finite, because there is a longest word. – Mark Bennet Mar 25 '15 at 19:39
  • @MarkBennet Haha tell me wich one then I give you a newly created one wich is longer. (There are tons of pre- and suffixes to chose from). – AlexR Mar 25 '15 at 19:41
  • @AlexR There is an official record. Of course you can make a longer one, but that may stretch the idea of an English word, but it is a game to play. However, if the universe is finite, then the number of possible words is ultimately finite, because there is a finite word length which exceeds the finite capacity of the universe to store information. – Mark Bennet Mar 25 '15 at 19:44
  • @MarkBennet Well that turns to a more philosophical / physical question. The mathematicians can deal with infinite amounts of information :P – AlexR Mar 25 '15 at 19:45

3 Answers3

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There are $26$ letters in the English language.

Consider each letter as one of the digits on base $27$:

  • $A=1$
  • $B=2$
  • $C=3$
  • $\dots$
  • $Z=26$

Then map each word to the corresponding integer on base $27$, for example:

$\text{BAGDAD}=217414_{27}=2\cdot27^5+1\cdot27^4+7\cdot27^3+4\cdot27^2+1\cdot27^1+4\cdot27^0$.

This mapping yields that the cardinality of your set is $\leq|\mathbb{N}|$, hence this set is countable.

barak manos
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The set $S_n$ of the English words with length $n$ is finite (this is almost obvious). So it's also countable. Why is it finite? The set $A_n$ of all sequences with length $n$ made up of latin characters is finite as it contains $26^n$ elements. Only some of these sequences are meaningful/actual English words. So $S_n \subset A_n$. So $S_n$ is also finite.

The set $T$ for which you have to prove that it is countable is:

$T = S_1 \cup S_2 \cup S_3 \cup ... $

Now you have this theorem:
"A countable union of countable sets is also countable"

Applying it you get that T is also countable.
Thus your statement has been proved.

peter.petrov
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2

The easiest way to show a set is countable is to provide a way of counting it - ie a rule to determine the position of any member within the set.

In this case we can start with all 1-letter "words", from a to z - there are 26 of these. Then we can continue with the two-letter words aa, ab, ... az, ba, bb .... zy, zz. There are 26^2 of these. And so on.

Any finite-length word will be assigned a unique position in this sequence, therefore the sequence is countable.

IanF1
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