# Confusion with open sets

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 $\mathbb{R}^2)$

• My response here addresses the same issue you're wondering about. Just pretend, in the first sentence of that post, that $G = [0,a)$ and $Y = [0,2)$. – Kaj Hansen Apr 5 '18 at 8:25
• Glad I could help :) Best of luck! – Kaj Hansen Apr 5 '18 at 8:31
• The key is your phrase "then there are points... toward the negative axis"... But an open ball means an open ball of the space, not an open ball of some larger space such as $\Bbb R^n$. When speaking of a space $X$ and a $set$ $Y\subset X,$ the phrase "open ball" will usually mean "open ball of $X$"... But when treating Y as a sub$space$ of $X$ we must distinguish an open ball of $X$ from an open ball of $Y$ – DanielWainfleet Apr 5 '18 at 19:16

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 $\mathbb R$.

• Just to reconfirm, for the open set $U$ open in $R$, the open ball $B_\epsilon (a)$ for every $a \in U$, it is not true that $\forall x \in R; x\in B_\epsilon (a) \subset U?$ – Aurora Borealis Apr 5 '18 at 8:35
• I do not fully understand what you are asking, but this might be useful. If you are working in a metric space (like $\mathbb R$) and $A\subseteq X$ then there are $2$ topologies that rise: 1) the subspace topology as described in my answer 2) the topology that is induced by the metric $d$ restricted to set $A$. Fortunately these two topologies coincide. If $d'$ denotes this restricted metric on $A$ then for $a\in A$ we have $B_d(a,r)=\{x\in X\mid d(a,x)<r\}\subseteq X$ and $B_{d'}(a,r)=\{x\in A\mid d'(a,x)=d(a,x)<r\}\subseteq A$. It is important to discern $B_d(a,r)$ and $B_{d'}(a,r)$. – drhab Apr 5 '18 at 8:44
• The ball $B(0,r)$ in my answer is actually the ball $B_{d'}(0,r)$ where $d'$ is the restriction of metric $d$ on on $[0,2)\times[0,2)$. It differs from $B_d(0,r)=\{x\in\mathbb R\mid |x|<r\}$ which contains negative elements and is not a subset of $[0,2)$. – drhab Apr 5 '18 at 8:50
• Ok thank you for the helpful explanation. Yes I think I am getting there. – Aurora Borealis Apr 6 '18 at 4:37