Why does $\{x\in \mathbb{R}: \lfloor y-x\rfloor<y-M, \mbox{ for all $y>M$ } \}=(M,\infty)$?

Question

Let $M$ be a fixed real number. During our research, we have obtained the following identities. $$ \{x\in \mathbb{R}: \lfloor y-x\rfloor<y-M, \mbox{ for all $y>M$ } \}=(M,\infty), $$ $$ \{x\in \mathbb{R}: \lfloor y-x\rfloor\leq y-M, \mbox{ for all $y\geq M$ }\}=[M,\infty). $$ It may be interesting to check them directly. Do you have a direct solution to prove it?

Observation. Let $X$ be the left side set of the first identity and $x\in X$. Putting $y=\lfloor M+1\rfloor$ we conclude that $x>M-1$. Also, if $x>M+1$, then $\lfloor y-x\rfloor\leq \lfloor y-M-1\rfloor<y-M$ and so $x\in X$. Thus $(M+1,\infty)\subseteq X\subseteq (M-1,\infty)$.

Answer 1

In both equalities, the inclusion $\supseteq$ results from $\lfloor y-x\rfloor\le y-x.$ Let us prove $\subseteq$ by contradiction.

Choose some integer $n>M-x,$ and let $y:=n+x>M.$

• Assume first $x\le M.$ Then, $\lfloor y-x\rfloor=n\ge n+x-M=y-M,$ which proves that $x$ doesn't belong to the set at the left of the 1st equality.
• Assume now $x<M.$ Similarly, $\lfloor y-x\rfloor=n>n+x-M=y-M,$ which proves that $x$ doesn't belong to the set at the left of the 2nd equality.

This proof by contradiction can be reformulated as a direct proof, with the same choice for $n$ and $y:$

• Assume first that $x$ belongs to the set at the left of the 1st equality. Then, $n=\lfloor y-x\rfloor<y-M=n+x-M,$ hence $x>M.$
• Similarly, assume now that $x$ belongs to the set at the left of the 2nd equality. Then, $n=\lfloor y-x\rfloor\le y-M=n+x-M,$ hence $x\ge M.$