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Let A be a nonempty set, such that $zin A$ and $sup(A) not in A$

(a) $sup(A setminus {z}) = sup(A)$

(b) generalise this to $sup(A setminus {z_1 z_2 z_3 dots z_n})=sup(A)$

Let us denote $A$ with $z$ removed by $B$

For the first part I know that removing a point from $A$ can only really lower the supremum, because we might remove the right "end point", so the only possibility for $sup(B)$ is that $sup(B) leq sup(A)$, we can also see this via this formal argument:

Let $y in A $, but $y neq z$, then we have that $y leq sup (B)$, but certainly we have that $yin B$ so now this is an upper bound for $B$, since $sup(B)$ is the lowest of bounds for $B$, we have:
$$ sup(B) leq sup(A)$$

I fail so see the other direction, the problem is if we start out with an arbitrary element of $B$ is that we cannot say anything about $z$, at least I don't see this.

For (b) we simply solve this by induction on $n$, here is a rough sketch, I think this approach is fine, but correct me if I am wrong:

Our base case will hold by question (a), our inductive hypothesis generalises this for arbitrary $k in mathbb{N} $ to $A setminus {z_1,z_2,dots,z_k}$ and then for our inductive step we first consider $A setminus {z_1,z_2,dots,z_k }=C $ this is useful notation to make things shorter.

We can now write that: $sup(Csetminus{z_{k+1}})=sup(C)$ by question $(a)$,

We now use our inductive hypothesis ($sup(C)=sup(A)$) and state that: $$sup(Asetminus{z_1,z_2,dots,z_k,z_{k+1}})= sup(Csetminus{z_{k+1}})=sup(C)=sup(A)$$
By the principle of mathematical induction the desired relation holds. $square$

Suppose that $sup(B)<sup(A)$. Then, take $xin A$ such that $sup(B)<x<sup(A)$ (We can do it by definition of $sup$ and because $sup(A)notin A$). Then, we have that $xnotin B$ hence $x$ must be $z$. But it is impossible because it would imply that $z=sup(A)$. Then, we have that $sup(A)=sup(B)$.