What is complement of $S$ in $\mathbb{R^3}$ I have two short questions, and I think they are totally related!
Question1: If I take set $S=\{(x,x+y,x+y): x,y∈\mathbb{R}\}$ then $(\mathbb R^3\backslash S)=?$
My attempt: $(\mathbb R^3\backslash S)=\mathbb{R^3}-S$
$= \{(x,y,z)∈\mathbb{R^3}: z<y \text{ or }z>y\}$ is this correct?
Question2:  where can I express $(\mathbb R^3\backslash S)$ as union of two disjoint open sets in $\mathbb{R^3}$? 
My attempt: yes we can, 
$(\mathbb R^3\backslash S)=\{(x,y,z)∈\mathbb{R^3}: z<y \text{ or }z>y\}$
$=\{(x,y,z)∈\mathbb{R^3}: z<y\} ∪ \{(x,y,z)∈\mathbb{R^3}:z>y\}$ is this is correct?
Please help me. my attempts are correct or not? and if yes, then in general how can we find complement of set in higher dimensions?
 A: You did exactly the right thing by including your own solution. I'm going to write elements of $S$ as $(s, u, u)$ where $u = s + t$, rather than in terms of $x$ and $y$, because that gives me the freedom to use $x, y, z$ for the first, second, third coordinates. This description is complete, because if I give you $(s, u, u)$, you can recover from it $t = u-s$, so every item of the form $(s, u, u)$ is in your $S$, and every item in $S$ has the form $(s, u, u)$. 
As you have observed, any point in $S$ has $y = z$, but $x$ can be anything. Therefore points with $y \ne z$ (and with any $x$-value at all) are not in $S$. So your description of the complement of $S$ as the union of those two sets is correct. 
Are the two sets disjoint? Clearly yes, for $y > z$ and $y < z$ are incompatible. 
Are the two sets open? Yes, each is an open half-space. That takes a little proving, but not much. The easiest proof I can see is to consider the map 
$$f: \Bbb R^3 \to \Bbb R^2 : (x, y, z) \mapsto y-z$$.
This map is evidently continuous. But your two sets (let's call them $P$ and $Q$) are simply
$$
P = f^{-1}( (0, \infty) )\\
Q = f^{-1}( (-\infty, 0) )
$$
where by $(0, \infty)$, I mean the open interval consisting of all positive reals. Because the preimage of an open set under a continuous map is always open, your set $P$ is open. A similar argument works for $Q$. 
Nice work!
One last thing: you've asked a second question (how can we find the complement of a set in higher dimensions?), which is generally frowned upon, but I'll give a quick answer here anyhow: in general, it's tough. Finding nice mappings between such complements and things we "understand" (like axis-aligned half-spaces, etc.) is often a big challenge. For instance, if you take an arc in 3-space, tie a (loose) knot in it, and then glue the ends together, you get something that topologists call a "knot". The complement of this knot turns out to sometimes be topologically quite complicated, and tools for simply describing knot-complements are themselves nontrivial. And that's for a single arc in 3-dimensions!
