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This Question comes from Measure Theory, Probability, and Stochastic Processes by Jean-François Le Gall ,page 72 Theorem 4.8

Background:Let $(E,d)$ be a metric space with metric $d$

Notation : (1)Let $A\subset E,x\in E$,define $$d(x,A):=\inf_{z\in A}d(x,z)$$(2)$$d(x,A)\wedge 1:=\min\{d(x,A),1\}$$(3)A function $f:E\to \mathbb R$ is said to be Lipschitz if$$\exists K\ge 0,\forall x,y\in E ,\quad|f(x)-f(y)|\leq K d(x,y).$$Question:Let $O$ be an open set in the metric space i find it hard to show the function $d(x,O^c)\wedge 1$ is Lipschitz,or more precisely:$$\exists L\ge0,\forall x,y\in X,|d(x,O^c)\wedge 1-d(y,O^c)\wedge 1|\le Ld(x,y)$$The condition that $O$ is an open set may be useful,but i do not know how to use it yet.

My attempt to the question:i know that$$|d(x,O^c)\wedge 1-d(y,O^c)\wedge 1|\le |d(x,O^c)-d(y,O^c)|$$Although $|d(x,z)-d(y,z)|\le d(x,y),\forall z\in O^c$can be showed by triangle inequality,i am stuck to show that$$|d(x,O^c)-d(y,O^c)|\le d(x,y)$$

Qo Ao
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  • Show that $d(x,O^{c})\le d(y,O^{c})+d(x,y)$ and $d(y,O^{c})\le d(x,O^{c})+d(x,y)$ – Kavi Rama Murthy Oct 07 '24 at 07:36
  • Does this answer your question? https://math.stackexchange.com/q/48850/42969 – Martin R Oct 07 '24 at 07:37
  • $x\mapsto d(x, A)$ is Lipschitz if $A$ is non-empty, if $A = \emptyset$ then $d(x, A) = \infty$ so this is a type of pathology. Moreover, minimum of Lipschitz functions is Lipschitz. And if $A = \emptyset$ then your function is simply the constant function which maps everything to $1$, so that by taking minimum you are avoiding the pathology of the empty set – Jakobian Oct 07 '24 at 07:44
  • Thank you!! I'v known how to solve my question! – Qo Ao Oct 07 '24 at 07:47

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