Looking through my lecture notes, I came across the notion that if a set $X\subset \mathbb{R}^n$ is compact and convex and $vol(X)=2^n$, then by choosing an $0<\epsilon <1$, then $X\subsetneq (1+\epsilon)X$, by convexity of $X$ which makes sense to me.
But if instead, we are given $vol(X)=1$, where $X$ is compact(but not necessarily convex), is there still reason to believe that for $0<\epsilon <1$, then $X\subsetneq (1+\epsilon)X$. Or can I choose some $\alpha$ instead where $X\subsetneq (1+\alpha)X$, where $\alpha$ has some specified bound.
EDIT: Would we still have the case that $vol((1+\epsilon)X)>vol(X)=1$, regardless of whether or not $(1+\epsilon)X$ contains $X$
The following regards the Edit made to the Question, about the "volume" of $(1+\epsilon)X$ in comparison to that of $X$.
Assume $X$ is Lebesgue measurable with "volume" $1$ in $\mathbb{R}^n$. Then for any $0 \le \epsilon$, the "dilation" $(1+\epsilon)X$ will be Lebesgue measurable with "volume" $(1+\epsilon)^n$.
This is one of the properties of Lebesgue measure. More generally if $T$ is a linear transformation on $\mathbb{R}^n$ (of which the dilation above is a special case), then $T(X)$ will again be measurable and have "volume" $|\det(T)|$.