Let consider the subspace $A = [0,1] \cup \{3\} \subseteq \mathbb{R}^1 $. We do now that the subspace topology on $A$ coincides with the induced metric topology on $\mathbb{R}^1$. (Usual Euclidean metric).
Now, this implies that $$\{3\} = B_{\mathbb{R}}(3,1) \cap A,$$ hence $\{3\}$ is open in $A$.
However, this implies $\exists \delta > 0$ s.t $B_{A}(3, \delta) \subseteq \{3\}$, and since we have found a open ball around $3$ that is contained in $A$, $3$ is an interior point of $A$, but this contradicts with what I have learned in my analysis courses (about the topology on $\mathbb{R}^n $), so what is the mistake that I'm doing in this argument ?
Edit:
Apparently, I wasn't realised that we define the concepts such as $closure, boundary, interior ...$ for subsets of a topological space, so after seeing this, the problems resolves itself.
The question can be closed.
$B_{A}(3,\delta)=\{x\in A: |x-3|<\delta\}$. For example, if $0<\delta<1/2$, then $B_{A}(3,\delta)=\{3\}$.