# Open set metric space definition in Topological space

Returning to a question after some time, I couldn't still quite figure out what is missing in my understanding.

At the heart, my question is this:

Given a subspace topology $\jmath$ on a subset A=$\left ( -1,\frac{-1}{2} \right ] \cup \left [ \frac{1}{2},1 \right )$ of the $\mathbb{R}$, we have a standard topology. A is a subspace of $\mathbb{R}$.

But when is it appropriate to use the metric space definition for the open set in order to determine whether A is an open set with respect to $\mathbb{R}$? One issue is that because A is not a basis for the standard topology by a certain theorem. So we are unable to argue via the definition of topology generated by basis that $A\subseteq \mathbb{R}$ is an open set with respect to $\mathbb{R}$.

I have been told that the only way is to use the definition of open set in the context of a metric space. However, a topological space is not necessarily a metric space.

Any clarification would be illuminating.

The metric space definition of open set still works: a subset of $A$ is open in $A$ if and only if it contains an open ball in $A$ about each of its points. For instance, in $A$ the open ball of radius $\frac14$ centred at $\frac12$ is
$$B_A\left(\frac12,\frac14\right)=\left\{y\in A:\left|y-\frac12\right|<\frac14\right\}=\left[\frac12,\frac34\right)\;,$$
so $\frac12$ is actually in the interior of $\left[\frac12,1\right)$ in the space $A$. Thus, $\left[\frac12,1\right)$ is open in $A$ not just as determined by the relative topology, but as determined by the metric open balls in $A$. You just have to realize that these balls are the same as the corresponding balls in $\Bbb R$ only when the corresponding balls in $\Bbb R$ are subsets of $A$.
• @Mathematicing: Yes. And that is the case, since the open balls in question are the open balls in $A$, not the open balls in $\Bbb R$. Notice the restriction to $y\in A$ in the definition of $B_A\left(\frac{1}2,\frac{1}4\right)$, for instance. Sep 8 '16 at 4:05