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?
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!