Let $(t,x)$ be in $(0,\infty)\times(0,1$) with the constraint $x+t<1$. Let $(a,b)$ be a subset of $(0,1)$. I want to know for which values of $t$ we have $${(x,x+t)}\cap{(a,b)}\neq\emptyset$$ or $${(x-t,x)}\cap{(a,b)}\neq\emptyset$$ with the contraint $x-t<1$.
Thank you.
Its strange that you let $t \in (0, \infty)$ and $x \in (0,1)$ but then restrict $t+x<1$. This means $t$ can't possibly be more than $1$. Also the restriction $x-t<1$ holds for all elements of this space since $t$ is positive and $x$ is less than $1$.
That said, if $x>b$ then $(x,x+t) \cap (a,b) = \emptyset$. If $x \in [a,b)$ then $(x,x+t) \cap (a,b)$ is never empty. Finally if $x<a$ then $(x,x+t) \cap (a,b)$ is non empty if and only if $x+t>a$.
The $(x-t,x)$ case is similar.