With some help earlier in the my other post, an open set $U $ in $\mathbb{R}^n$ is considered open iff it contains $n$-dimensional open balls at each of its points in $U.$ But if we have an open set $U$ in $R$ where $R=[0,2),$ can the open set for this $R$ be $[0,a)$ where $a<2?$
Because if $[0,a)$ was an open set inside $[0,2)$, then we cannot find an open ball for every point inside $[0,a)$ as if have some open ball at the point $0$, then there are points which lie outside of $0$ towards the negative axis, and that is no longer inside the set $U$. So where am I going wrong with the understanding open sets?
Also why is the open square $(a,b)\times(c,d)$ open in the upper half plane in $\mathbb{R}^2_+$, AND $\mathbb{R}^2?$ Furthermore, for a partially closed square, $(a,b)\times [c,d)$ in $\mathbb{R}^2$, why is this closed in $\mathbb{R}^2_+$ but not closed in $\mathbb{R}^2?$ $(\mathbb{R}^2_+$ is the upper-half plane in the Euclidean space $\mathbb{R}^2)$
If you are wondering whether $[0,a)$ an open set in $[0,2)$ then you do not have to be concerned about points that are not elements of $[0,2)$.
Working in $[0,2)$ the open ball $B(0,r)$ is defined as $\{x\in[0,2)\mid |x|<r\}$.
For $r$ small enough we have $0\in B(0,r)\subseteq[0,a)$.
If $X$ is some topological space and $A$ is a subset then $A$ inherits the so-called subspace topology.
This comes to: a set $U\subseteq A$ is open in $A$ if it can be written as intersection $A\cap V$ where $V$ is open in the original topology on $X$.
Note that $[0,a)=[0,2)\cap(-\infty,a)$ where $(-\infty,a)$ is open in $\mathbb R$.