Let $ A\subset\mathbb{R} $ a bounded set, $ \mathbb{N} = \{0,1,2,...\} $ and $$ B:= \left\{\frac{n}{n+1} - a \mid a \in A, n \in \mathbb{N}\right\}$$ Give inf B and sup B in terms of inf A and sup A. Also prove these statements.

So I thought $\sup B = 1 - \inf A.$

Proof for this: $\dfrac{n}{n+1}$ is bounded from above by 1 and $\lim_{n \rightarrow \infty} \dfrac{n}{n+1}=1.$

Further $\forall a\in A: -1+a\geq-1+\inf A$, so $\forall a\in A: 1-a\leq1-\inf A$. So $\sup B = 1 - inf A$.

For $\inf B$, I thought $\inf B = - \sup A$.

Proof for this: $\dfrac{n}{n+1}\geq0$ $\forall n\in\mathbb{N}$. $\forall a\in A:a\leq \sup A$, so $\forall a\in A:-a\geq -\sup A$. So $\inf B = 1 - \sup A$.

Can anyone tell me how formally correct this proof is and what I should improve in this proof.


Your proof looks correct, but I might show the intermediate step like this:

Let $C=\left\{\frac{n}{n+1} \;\Big\vert\; n\in\mathbb{N}\right\}$ so that $$B = \left\{c-a \mid c \in C, a \in A\right\}.$$ Now, $$\sup B = \sup C - \inf A = 1 - \inf A, $$ and $$\inf B = \inf C - \sup A = 0 - \sup A. $$ Ensure you prove what $\sup C$ and $\inf C$ are correctly and you should be done. (In this case $\inf C=\min C$, so that's easy. $\sup C$ takes only a little bit of effort.)


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