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What does the set $C(x,t,r)=(y,s)|$such that $|x-y| \le r ; s\in(t-s^2,t) $ represent . I am not sure if the question is well posed . Does it represent cylinder ?

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You don't say what $x$, $y$, $r$, $s$ and $t$ are. Also, is $(y,s)$ supposed to be a pair or an open interval (too bad the standard notation doesn't distinguish both)?

I'll assume that $r$, $s$ and $t$ are intended to be real numbers (for $r$, there's not much else which would make sense), and $x$ and $y$ are members of some real or complex vector space (I assume that they are not intended to be just real numbers, because otherwise they probably would have been named with letters adjacent to $r$, $s$, $t$, e.g. $p$ and $q$, or $u$ and $v$). Also I assume that $(y,s)$ is meant to be a pair, while $(t-s^2,t)$ means an open interval in $\mathbb{R}$.

$\left|x-y\right|\le r$ just means that $y$ is a member of the closed ball of radius $r$ with center $x$ (let's denote it as $B_r(x)$ for future reference). $s\in(t-s^2,t)$ means $t-s^2. Now $t-s^2 can be rewritten as $t, which for $t<-\frac14$ is always true, otherwise only if either $s>-\frac12+\sqrt{t+\frac14}$ or $s<-\frac12-\sqrt{t+\frac14}$. Now $-\frac12+\sqrt{t+\frac14}\le t$ for all $t\ge -\frac14$, with equality only for $t=0$. Therefore for $t<-\frac14$, the second condition just means $s\in(-\infty,t)$, for $t>\frac14$ but $t\neq 0$ it means $s\in(-\infty,-\frac12-\sqrt{t+\frac14})\cup(-\frac12+t\frac14,t)$ and for $t=0$ it means $s\in(-\infty,-\frac12-\sqrt{t+\frac14})$.

Therefore the described set is $C(x,r,t) = \cases{ B_r(x)\times(-\infty,t) & for $t<\frac14$\\ B_r(x)\times\left((-\infty,-\frac12-\sqrt{t+\frac14})\cup(-\frac12+\sqrt{t+\frac14},t)\right) & for $t\ge\frac14\wedge t\neq 0$\\ B_r(x)\times(-\infty,-\frac12-\sqrt{t+\frac14}) & for $t=0$}$

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I think i got it . It represents a cylinder with top center$(x,t) $ with the height of the cylinder $r^2. $