371: For Metric Space, Difference of Distances of 2 Points from Subset Is Equal to or Less Than Distance Between Points

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A description/proof of that for metric space, difference of distances of 2 points from subset is equal to or less than distance between points

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Topics

About: metric space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of metric space.
• The reader knows a definition of distance between 2 points on metric space.
• The reader knows a definition of distance between point and subset on metric space.

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Target Context

• The reader will have a description and a proof of the proposition that for any metric space, any subset, and any 2 points, the difference of the distances of the points from the subset is equal to or less than the distance between the points.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any metric space, $M$, any subset, $S\subseteq M$, and any points, $p_1,p_2\in M$, $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$ where $d(\bullet ,\bullet )$ denotes the distance between the arguments.

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2: Proof

Suppose that $p_1,p_2\notin S$. Take any point, $p_2^{\prime }\in S$. By the definition of distance between 2 points, $d(p_1,p_2^{\prime })\le d(p_1,p_2)+d(p_2,p_2^{\prime })$. By the definition of distance between point and subset, $d(p_1,S)\le d(p_1,p_2^{\prime })$ and $d(p_2,p_2^{\prime })=d(p_2,S)+ϵ$ where $ϵ\ge 0$ can be made any small positive. So, $d(p_1,S)\le d(p_1,p_2)+d(p_2,S)+ϵ$. By taking the limit by $ϵ\to 0$, $d(p_1,S)\le d(p_1,p_2)+d(p_2,S)$, so, $d(p_1,S)-d(p_2,S)\le d(p_1,p_2)$. By the symmetricalness, also $d(p_2,S)-d(p_1,S)\le d(p_1,p_2)$, so $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$.

Suppose that $p_1\notin S$ and $p_2\in S$. By the definition of distance between point and subset, $d(p_1,S)\le d(p_1,p_2)$. As $d(p_2,S)=0$, $d(p_1,S)-d(p_2,S)\le d(p_1,p_2)$. As the left side is non-negative, $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$.

Suppose that $p_1\in S$ and $p_2\notin S$. By the symmetricalness, $|d(p_2,S)-d(p_1,S)|\le d(p_2,p_1)$, which implies $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$.

Suppose that $p_1,p_2\in S$. As $d(p_1,S)=d(p_2,S)=0$, $|d(p_1,S)-d(p_2,S)|\le d(p_1,p_2)$.

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References

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