Subsets and Greatest Integer Function

For $\lfloor x \rfloor = n$ where $n \in \{3, 4, 5\}$, we have $n \le x < n+1$. Then $\lfloor x \rfloor^2 = n^2$. The condition $\lfloor x^2 \rfloor = n^2$ implies: $n^2 \le x^2 < n^2 + 1 \implies n \le x < \sqrt{n^2 + 1}$

• For $n = 3$: $x \in [3, \sqrt{10})$
• For $n = 4$: $x \in [4, \sqrt{17})$
• For $n = 5$: $x \in [5, \sqrt{26})$
• For $x = 6$: $\lfloor 36 \rfloor = 6^2 = 36$, so $x = 6$ is included.

Hence $S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup \{6\}$. The set $(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}$ is a subset of $S$.