All elements of a set I am trying to solve the following problem;
Write all elements of the following set: $ A=\left \{ x\in\mathbb{R}; \sqrt{8-t+\sqrt{2-t}}\in\mathbb{R}, t\in\mathbb{R} \right \}$ .
My assumption is that the solution is $\mathbb{R}$ and we don't need to solve when are the square roots defined, because of the $x$. Am I correct?
Thanks
 A: The set $A$ is not properly defined. I am interpreting $A$ as the range of $\sqrt {8-t+\sqrt {2-t}}$.
$\sqrt {8-t+\sqrt {2-t}}$ is defined only when $t \leq 2$. This function is continuous and decreasing. It tends to $\infty$ as $ x \to -\infty$ and the value at $x=2$ is $\sqrt 6$. Hence the answer is $[\sqrt 6,\infty)$. 
A: I believe all the other answers misinterpret the question. The answer $A=[\sqrt{6},+\infty)$ would be correct if it was phrased as $A=\{x\in\mathbb{R}| x=\sqrt{8-t+\sqrt{2-t}}, t\in\mathbb{R}\}$. I am interpreting the question as "What is $A=\{x\in\mathbb{R}| \sqrt{8-t+\sqrt{2-t}}\in\mathbb{R}\quad \forall t\in\mathbb{R}\}$". Notice the quantifier at the end, which is often omitted in such cases.
I think the answer is the empty set. I think that the point of the problem is to teach you that things like $A=\{x\in B| 0=1\}$ construct an empty set. Your $A$ consists of $x$ such that for all $t$ $\sqrt{8-t+\sqrt{2-t}}$ is well-defined. Since it is always undefined for $t>2$ (or, even better, defined as a complex number with positive imaginary part), no $x$ satisfies this condition and $A$ turns out to be empty.
Basically, when you have a set defined as $A=\{x\in B| \phi\}$, where $\phi$ is some statement not dependent on $x$, then $A=B$ ($\phi$ is True), or $A=\varnothing$.
