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Say we want to show that the interior of a set $A$ is open.

If $x \in Int(A)$, then there exists an open ball $B_r(x) \subseteq A$.

Since $B_r(x)$ is open, $y \in B_r(x)$ also has an open ball $B_s(y) \subseteq B_r(x) \subseteq A$, so $y \in Int(A)$.

Now, somehow we have to show that the ball $B_r(x) \subseteq Int(A)$, and that would complete the proof.

All of the proofs I read say this is obvious, but I don't see how $B_r(x) \subseteq Int(A)$ immediately follows here.

ijuneja
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mr eyeglasses
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    You have, for an arbitrary $y\in B_r(x)$, shown that there is a ball $B_s(y)$ contained in $A$ and hence $y\in \operatorname{Int} A$. That means for all $y\in B_r(x)$ you have $y\in \operatorname{Int} A$, and that's by definition $B_r(x) \subseteq \operatorname{Int} A$. – Daniel Fischer Feb 25 '15 at 14:06
  • You just showed that $y \in B_r(x) \implies y \in Int(A)$. So $B_r(x)\subseteq Int(A)$ – Reznick Sep 12 '17 at 18:45
  • Are you saying the argument is circular? @Daniel Fischer – Plotinus Aug 20 '21 at 13:31
  • @Plotinus No, I don't think Daniel Fischer is saying the OP's argument is circular. Rather , I believe Daniel is bringing attention to the fact that the $y$ in $y\in B_r(x)$ can be chosen arbitrarily, and thus the OP showed, without realizing, that $ \forall y \left( y \in B_r(x) \implies y \in Int(A)\right) $. This is equivalent to $B_r(x)\subseteq Int(A)$, by definition of subset , viz., $A \subseteq B \Leftrightarrow \forall x (x \in A \implies x \in B)$. – john Jun 27 '23 at 05:50

2 Answers2

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$$\text{Int}(A)=\{x\in X\mid \exists r>0: B_r(x)\subset A\}.$$

Therefore $$\forall x\in \text{Int}(A), \exists r_x>0: B_r(x)\subset A,$$ and thus $$\text{Int}(A)=\bigcup_{x\in \text{Int}A}B_{r_x}(x).$$

We finally conclude that $\text{Int}(A)$ is open.

Surb
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  • How does $x \in \text{Int}(A), B_r(x) \subset A$ imply $$\text{Int}(A) = \bigcup_{x\in\text{Int}(A)} B_r(x)$$ ? – feynhat Feb 18 '18 at 09:17
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    @feynhat well as $B_r(x)$ is open, we have $B_r(x)\subset int(A)$ and so $int(A) = \cup_{x\in int(A)} x\subset \cup_{x\in int(A)} B_r(x) \subset int(A)$. – Surb Feb 20 '18 at 06:18
  • All in one def:Int(A)=$\bigcup$ {U $\subset$ X : U is open and U $\subset$ A } – Plotinus Aug 20 '21 at 13:27
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If $ x \in int(E) $ then $ \exists r >0 $ , $ B_r(x) \subset E $

Now $ \forall y \in B_r(x) \subset E $, $ \exists r_y = r - d(x,y) $, $ \ni B_{r_y} \subset B_r (x) \subset E $ Hence $ y \in int(E) $

Hence $ B_r(x) \subset int(E) $ or x is interior point of int(E)

niraj panakhaniya
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