proving that the set of all english words is countble. 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 !
 A: 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.
A: 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.
A: 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.
