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Let $A$ be an arbitrary metric space.

$$A^\circ= \{x\mid x \textrm{ is an interior point of } A\}$$

Show that $A^\circ$ is an open set.

I tried to make a contradiction argument by arguing that if $A^O$ was not open it must have a boundary point $a^O\in A^O$ (I assume the sets are not empty or the universe set) and then by construction of $A^O$ there must exist two balls $B(a^O,r/2) \subseteq B(a^O,r)\subseteq A$ which has at least one element $x$ from $A\setminus A^O$. But then $B(x,r/2)\subseteq A$ is an open ball in $A$ and $x$ should be in $A^O$ and not in $A\setminus A^O$. $\Rightarrow\Leftarrow\quad$

Is this a valid proof or just rambling?

Thanks in advance

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    By "arbitrary set" do you mean "arbitrary subset of $\mathbb{R}^n$ for some $n>0$"? Or something else? – 2017-01-22
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    What is your definition of an `open set`? – 2017-01-22
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    A subset A of a metric space is open if it does not contain any of its boundary points. A point x is a boundary point in A if the ball B(x,r) for all r>0 contains points in A and its compliment $A^C$. Do I have to restrict myself to a subset of $ℝ^n$, i was thinking about any metric space. – 2017-01-22
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    Ah, so it's a subset of a metric space! You should edit this information into your question. (Also, if you don't want to restrict yourself to $\mathbb R^n$, why did you add the (real-analysis) tag?) – 2017-01-22

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It seems like you're on the right track, but consider a simpler proof:

Any element $x \in A^{\circ}$ has an open ball $B(x,r_x)$ around it contained in $A$. Thus the union of all such balls is open. Clearly we have $A^{\circ} \subset \cup_{x} B(x,r_x)$. Conversely, if $y \in \cup_{x} B(x,r_x)$, then $y \in B(x,r_x)$ for some $x$. Since the ball is open, there exists a smaller ball around $y$ also contained in $A$. It follows that $A^{\circ} = \cup_{x} B(x,r_x)$, so $A^{\circ}$ is open.