Compact set in complex plane If $(a, b) \subset \mathbb{R} \subset \mathbb{C}$, and it can be shown that $(a, b)$ is an open subset of $\mathbb{R} $ yet it is not an open subset of $\mathbb{C} $.If $[a, b] \subset \mathbb{R}$ is compact, how would you show $[a, b]$ is compact in $\mathbb{C}$? Whats the difference in the proof?
 A: The topology of $\mathbb{C}$ is the same or $\mathbb{R}^2$, hence a subset of $\mathbb{C}$ is compact iff it'a closed and bounded. Indeed $[a,b] \times {0} $ is closed and bounded, hence compact.
A: If we inject $[a,b]$ into the complex plane (which is also viewed as ${\bf{R}}^{2}$), then it becomes $[a,b]\times\{0\}$. The latter set is closed and bounded.
A: As I understand it, you have two questions.  
Question 1: Demonstrating that $(a,b)$ is not an open subset of $\mathbb{C}$. First $(a,b)$ is not a subset of $\mathbb{C}$ as $\mathbb{C}=\mathbb{R}\times\mathbb{R}$.  But it is clear what you mean, you are thinking about those real numbers as living in the complex field. As a set of real numbers we have $(a,b):=\{x\in\mathbb{R}: a<x<b\}$, and for an $x\in\mathbb{R}$ to be thought of as an element of of the complex field $\mathbb{C}$ we write $z=x+i0$, or represent it as the ordered pair $z=(x,0)$. So naturally when you say $(a,b)$ is a subset of $\mathbb{C}$ you are really referring to a line segment on the horizontal axis $(a,b)\times\{0\}:=\{(x,y)\in\mathbb{C}: a<x<b, y=0\}$. This subset of $\mathbb{C}$ is not open because there exists an element in the set that is not interior to the set, in particular the midpoint $(\frac{a+b}{2},0)$ is not interior because every open neighborhood is not contained in the set. In fact, all the elements of the set are not interior, instead they are all boundary points. Note that set is not closed either because it doesn't contain all its boundary points, in particular the boundary points $(a,0)$ and $(b,0)$ are not in the set.
Question 2: Demonstrating that $[a,b]$ is compact in $\mathbb{C}$.  Following the same discussion we had in Question 1, it is safe to assume that you are referring to the set $[a,b]\times\{0\}\subset\mathbb{C}$. This set is compact because it is closed and bounded (which is equivalent to compact). It is bounded because it can be contained in a neighborhood of the origin, in particular a neighborhood of radius $\delta>\max\{|a|,|b|\}$.  It is closed because it contains all of its boundary points.
A: If A is a sub$space$  of the space $B$ then $A$ is compact iff $A$ is "compact in $B$".
If $A$ is a subspace of $B$ and $B$ is a subspace of $C$ then $A$ is a subspace of $C.$
Apply this with $A=[a,b], B=\Bbb R,$ and $C=\Bbb C.$
In other words, the topology on $[a,b]$ as a subspace of $\Bbb R,$ which is a compact topology, is the same topology that $[a,b]$ has as a subspace of any space $C$ such that $\Bbb R$ is a subspace of $C.$   
