I recently solved a problem in which I used the following fact.
If there are exactly $n$ integers in the interval $[x,y]$ ($x,y\in\mathbb Z$) and $a_1,a_2,\ldots,a_n$ are integers of that interval and furthermore, $a_1<a_2<\cdots<a_n$ then necessarily $a_1=x,$ $a_2=x+1,$ $\ldots,$ $a_n=y.$
My question is, how can I rigorously prove this fact?
I think that the pigeonhole principle is enough, for if $a_k\neq x+k-1$ for some $k,$ the pigeonhole principle would imply that there are $i$ and $j,$ $i\neq j$ such that $a_i=a_j,$ which is a contradiction. However, I am not completely sure that this is rigorous enough.
I appreciate your help!
Yes, you are right. The pigeonhole principle is enough. First, you can change your chain of inequalities to the statement that all integers $a_i$ are pairwise disjoint i.e. for each $i \neq j$, $a_i \neq a_j$. Then you know that the set of all $a_i$ is of size $n$. But it is a subset of the set $[x, y]$. So it must be the same set. Re-applying the ordering easily gives you $a_1 = x$, $a_2 = x + 1$, etc. and finally $a_n = y$.
Edit: I realise that I didn't really explicitly explain my use of the pigeonhole principle in the above. I'll explain now using the pigeonhole principle. Consider the set $[x, y]$ containing $n$ integers. Now consider the numbers $a_1, a_2, \dots, a_n$. These numbers can be viewed as a function from the set $\{1, 2, \dots, n\}$ into the set $[x, y]$. Now, imagine the numbers $\{1, 2, \dots, n\}$ are pigeons, and the numbers in $[x, y]$ are holes. The function from $\{1, 2, \dots, n\}$ to $[x, y]$ puts each pigeon into some hole. But note that clearly by the chain of inequalities, and the transitivity of the less than relation $<$, if $i < j$ then $a_i < a_j$ and so $a_i \neq a_j$. So no two pigeons can go in the same hole. Now, the pigeonhole principle says if there are $n$ pigeons, and $n$ holes, and no two pigeons can go in the same hole, there is exactly one pigeon per hole. Which means for each integer $u \in [x, y]$, there is some $i \in \{1, 2, \dots, n\}$ such that $a_i = u$.
For the ordering you need a slightly stronger argument. You want to show that $a_1 = x, a_2 = x + 1, \dots, a_n = y\;$ i.e. $a_i = x + i - 1$ for all $i \in \{1, 2, \dots, n\}$. Well, suppose this wasn't the case. Then there would be some smallest $i$ such that $a_i \neq x + i - 1$. So we would have $a_i \geq x + i$, because $i$ is the smallest such $i$. But since we've already shown that there is some $j$ such that $a_j = x + i - 1$ and again because $i$ is the smallest such $i$, we know that $j > i$. But we also have from the previous inequalities $a_j < a_i$, which contradicts the given ordering $a_1 < a_2 < \dots < a_n$. So we can conclude that $a_i = x + i - 1$ for all $i$.
Note this is extremely verbose, and one or two lines like your question or my original answer would suffice in any standard proof. But because we're dissecting a proof step here, I though I'd be more explicit and rigorous with the details. Maybe I've spent too long writing machine proofs :)