In other words, you have an ordered finite list of numbers which is sorted ascendingly (without loss of generality), for example [1,2,5,7,10,12]. Remove elements from the list, the list is still sorted. Prove this mathematically.
I know this is intuitive but how to prove it mathematically.
I tried induction but I struggled with mathematical representation of a sorted list (how to denote a sorted list mathematically).
Any help please?
Suppose
$$a_1, a_2, \ldots, a_n$$
is sorted. This means
$$i \geq j \implies a_i \geq a_j \tag{1}$$
Now we remove $a_k$.
$$a_1, \ldots, a_{k-1}, a_{k+1}, \ldots, a_n$$
Now show that this new list satisfies the property of a sorted array. If $i \geq j$ and $i,j \neq k$, then it is still true that $a_i \geq a_j$ from $(1)$. Thus the array is still sorted.