topology: is $(1/2,1]$ open in $[-1,1]$? If $Y = [-1,1]$ is subspace of $\Bbb R$, is $(1/2,1]$ open in $Y$? 
And what about in $\Bbb R$? 
I know $(1/2,1)$ is open in both $Y$ and $\Bbb R$ but I am confused when it is half closed and half open.
 A: The subspace topology on $Y\subseteq X$ is defined by: 
$\tau_Y:=\{U\cap Y| U\,\,\text{open in}\,\,X\}$
To decide if $(\tfrac12, 1]$ is open in $Y$ (with regards to $\tau_Y$), we have to give an open set U in $\mathbb{R}$, with $U\cap [-1,1]=(\tfrac12, 1]$.
And indeed we can give this easily:
Because $(\tfrac12, 2)$ is open in $\mathbb{R}$ and $(\tfrac12, 2)\cap [-1,1]=(\tfrac12, 1]$
A: 
Definition: Let $(X, \tau_X)$ be a topological space. Let $Y \subset X$ and define $\tau_Y$ as the collection of all sets of the form $Y \cap U$ where $U$ is in $\tau_X$. In other words,
  $$
\tau_Y \equiv \{ Y \cap U \colon U \in \tau_X \}.
$$
  Then, $\tau_Y$ is called the subspace topology of $Y$ with respect to $(X, \tau_X)$.

In your case, $X = \mathbb{R}$, $\tau_X$ is the standard topology generated by the absolute value, and $Y = [-1, 1]$.
You should be able to use the above definition to answer your questions.
A: Sure $(\frac{1}{2},1]$ open in $Y$, e.g. because it's the open ball 
$B_{d_Y}(1,\frac{1}{2})  = \{x \in Y: d_Y(x,1) < \frac{1}{2}\}$, where $d_Y$ is the restriction of the Euclidean metric to $Y$, and open balls are by definition open in the metric topology induced on a set.
