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I've been told that $$d(x,A) = \inf_{y \in A} |x-y|$$ is uniformly continuous, but I don't understand why? Is there a short proof of this statement or is this a slightly deeper result? This was a result discussed in my analysis lecture.

Michael Hardy
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dalastboss
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    I think $\delta=\varepsilon$ will work. The main tool of the proof, if I'm not mistaken, will be the triangle inequality. So it's not only uniformly continuous, but Lipschitz-continuous. – Michael Hardy Dec 11 '14 at 19:39
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    I changed $inf_{y\in A}$ to $\displaystyle\inf_{y\in A}$. That is standard usage. – Michael Hardy Dec 11 '14 at 19:42

2 Answers2

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The reason $d(x,A)$ is a uniformly continuous function is that $|d(x,A) - d(y,A)| \le |x -y|$ for all $x$ and $y$. Let $x$ and $y$ be given. For every $z\in A$, $$d(x,A) \le |x-z| \le |x-y| + |y-z|.$$ Thus $d(x,A) - |x-y|$ is a lower bound for the set $\{|y-z|:z\in A\}$. Hence $d(x,A) - |x - y| \le d(y,A)$, i.e., $d(x,A) - d(y,A) \le |x - y|$. A similar argument shows that $d(y,A) - d(x,A) \le |x - y|$. Hence $|d(x,A) - d(y,A)| \le |x - y|$.

kobe
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$$ d(w,y)\le d(w,x)+d(x,y) $$ $$ \inf_{y\in A} d(w,y) \overset{\text{?}} \le \inf_{y\in A}(d(w,x)+d(x,y)) \overset{\text{?}}= d(w,x)+\inf_{y\in A} d(x,y) $$

Can you then show that if $d(w,x)<\varepsilon$ then $|d(w,A)-d(x,A)|<\varepsilon$?

Michael Hardy
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