Don't understand proof that interior of a set is open

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

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.

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General Topology

Real Analysis

Elementary Set Theory

Solution Verification

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