Let $A=(4,10] \cup \{1, 14, 27 \}$. Find, with proof, the interior point of $A$; that is, $A^o$, in $\Bbb R$ with the usual metric. Let $A=(4,10] \cup \{1, 14, 27 \}$. Find, with proof, the interior point of $A$; that is, $A^0$, in $\Bbb R$ with the usual metric.
EDIT
Attempt: I claimed that $A^0 = (4,10)$. Here's what I tried.

*

*Let $x \in (4,10)$ be arbitrary. Choose $r = \min\{x-4, 10-x\}>0$. The goal is to show that $B(x,r) \subseteq A$. Now, let $y \in B(x,r)$ be arbitrary. We have
\begin{align*}
|y-x|<r &\iff -r < y-x < r \\
&\iff x-r < y < x+r \\
&\iff 4 = x-(x-4) \le x-r < y < x+r \le x+(10-x) = 10.
\end{align*}
Hence, $y \in A$, and so, $B(x,r) \subseteq A$. Therefore, $(4,10) \subseteq A^0$.

*For $x=1$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=1+r_1$, where $r_1=\min\left\{\frac 12, \frac r2 \right\}>0$, we have $a \in B(1,r)$, but $a \notin A$.

*For $x=10$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=10+r_1$, where $r_1=\min\left\{\frac 12, \frac r2 \right\}>0$, we have $a \in B(10,r)$, but $a \notin A$.

*For $x=14$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=14+r_1$, where $r_1=\min\left\{\frac 12, \frac r2 \right\}>0$, we have $a \in B(14,r)$, but $a \notin A$.

*For $x=27$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=27+r_1$, where $r_1=\min\left\{\frac 12, \frac r2 \right\}>0$, we have $a \in B(27,r)$, but $a \notin A$.

*Notice that $A^0 \subseteq A$; indeed, because $x \in B(x,r)$ and $B(x,r) \subseteq A$, then $x \in A$. So, by taking the contraposition, if $x \notin A$, then $x \notin A^o$.

Thus, $(4,10) = A^0$, as claimed.
Does the above correct? Thanks in advanced.
 A: Your proof is not only correct, but also very clearly written. One minor thing you can do to shorten it. Because points 2,3,4,5 all basically use the same idea, I wouldn't write them as separate points. Instead, I would replace



*For $x=1$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=1+r_1$, where $r_1=\min\left\{\frac 12, \frac r2
> \right\}>0$, we have $a \in B(1,r)$, but $a \notin A$.

*For $x=10$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=10+r_1$, where $r_1=\min\left\{\frac 12, \frac r2
> \right\}>0$, we have $a \in B(10,r)$, but $a \notin A$.

*For $x=14$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=14+r_1$, where $r_1=\min\left\{\frac 12, \frac r2
> \right\}>0$, we have $a \in B(14,r)$, but $a \notin A$.

*For $x=27$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=27+r_1$, where $r_1=\min\left\{\frac 12, \frac r2
> \right\}>0$, we have $a \in B(27,r)$, but $a \notin A$.


with a much shorter and just as clear:



*For $x\in\{1, 10, 14, 27\}$, given any $r>0$, we always have $B(x,r) \nsubseteq A$. Indeed, by taking $a=x+\min\{\frac12, \frac r2\}$, we have $a \in B(x,r)$, but $a \notin A$.


