$Interior of union of two sets with empty interior$

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$Interior of union of two sets with empty interior$

Let $X$ and $Y$ be subsets of $\mathbb{R}^n$. If $\text{int}(X)$ and $\text{int}(Y)$ are empty sets and $X$ is closed, then $\text{int}(X \cup Y)$ is an empty set.

Real Analysis

Mona Vinci

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Let $z\in int(X\cup Y)$. Then every ball around $z$ must contain points from both $X$ and $Y$ since otherwise some ball around $z$ will be completely inside say $Y$, contradicting the fact that $Y$ has empty interior. Consider the balls of radius $1/n$ around $z$. Each of these balls contains a point in $X$ which we call $x_n$. Then $x_n$ converges to $z$ as $$|x_n -z|\le 1/n\to 0 $$ So $x_n \to z$ and thus $z\in X$ as $X$ is closed. Thus we showed that $$int(X\cup Y)\subset X $$But $int(X\cup Y)$ is an open set, so its interior is itself. Hence $$ int(X\cup Y)=int(int(X\cup Y))\subset int(X)=\emptyset$$ So $int(X\cup Y)=\emptyset$.

Erdos

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$Interior of union of two sets with empty interior$

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