Why is [0,1/2) open in [0, 1]? With the usual metric in $\mathbb{R}$, there is no open interval around $0$ that is completely contained in $[0,1/2)$. However, I have come across the fact that "$[0,1/2)$ is open in $[0,1]$" in a proof I am going through. It seems to directly contradict the definition of openness. A rationale behind this would be appreciated.
 A: You have to think in terms of the subspace topology. Let $\tau$ be the usual topology on $\mathbb R$ and $S=[0,1]$ and $A=[0,1/2)$. Then $A = S\cap U$, where $U$ is for example $(-1/2,1/2)$ (which is an element of $\tau$), so $A$ is open in the subspace topology, which is defined as $\tau_S = \{S\cap U : U\in \tau\}$.
A: An alternate but equivalently enlightening approach to those given by the other answers is to consider the perspective of boundary points.
A point $x \in X$ is called a boundary point of the subset $A\subset X$ if every open neighbourhood around $x$ (in $X$) contains at least some points from $A$ and some points from $A$-complement. 
(i.e. $x\in X$ is a boundary point of $A$ if there exists an open neighbourhood $U \subset X$ around $x$ such that $U\cap A \neq \emptyset$ and $U\cap A^C \neq \emptyset$)
A subset is closed iff it contains all of its boundary points, and is open iff it contains none of its boundary points. (It is neither open nor closed if it contains some but not all boundary points. The empty set and entire set $X$ are both open and closed since they both have no boundary points and hence vacuously both contain all of their boundary points and also contain none of their boundary points.)
The only boundary point of $[0,\frac12)\subset [0,1]$ is $\frac12$. 
(No points on the interior of $[0,\frac12)$ can be contained in the boundary since one could always take an arbitrarily small neighbourhood contained entirely within $A$, and $0$ is not a boundary point for the same reason, intuitively since it’s not “next to” any points in $A$-complement)
Then since $[0,\frac12)$ does not contain its only boundary point, it is open. 
