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Let $F$ be a closed set of $ \mathbb{R} $ whose complement has finite measure. Let $\delta(x) = d (x, F) =\inf \{ |x - z| \mid z \in F\}$.

Prove $ \delta$ continuous by proving $| \delta(x) - \delta(y) | \leq |x - y|$

I appreciate any kind of hint. Thanks

2 Answers 2

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Let $x, y\in\mathbb{R}$. For any $\epsilon>0$, by definition of $\delta(y)$, there exists $z\in F$ such that $|y-z|\leq\delta(y)+\epsilon.$ As Ashok said, we have the triangle inequality $|x-z|\le |x-y|+|y-z|.$ Combining these two inequalities, we have $|x-z|\le |x-y|+\delta(y)+\epsilon.$ By definiton of $\delta(x)$, we have $\delta(x)\leq |x-z|$ since $z\in F$. Hence, we have $\delta(x)\le |x-y|+\delta(y)+\epsilon.$ Since $\epsilon>0$ is arbitrary, we have $\delta(x)\le |x-y|+\delta(y)$, or $\delta(x)-\delta(y)\le |x-y|.$ By symmetry, we also have $\delta(y)-\delta(x)\le |y-x|.$ Therefore, we obtain $| \delta(x) - \delta(y) | \leq |x - y|.$

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I don't think we need $F$ to be a closed set whose complement has finite measure etc. The result holds for any nonempty $F$.

Hint: $|x-z|\le |x-y|+|y-z|$ and take $\inf$ over $z\in F$.