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I'm at the beginning of learning metric spaces and I've been given a fundamental problem.

Let $\left ( X,d \right )$ is metric space. Let $Y\subset X$. For $x,y\in Y$ we put $d^{*}\left ( x,y \right )=d\left ( x,y \right )$. Prove that function $d^{*}:YxY\rightarrow \mathbb{R} $ is metric on Y.

My solution: I need to verify three metric properties:

  1. $d(x,y)\geqslant 0,\hspace{0.2cm} \forall x,y\in X\hspace{0.2cm}\text{and}\hspace{0.2cm}d(x,y)=0\hspace{0.2cm}\text{iff} \hspace{0.2cm}x=y\\$
  2. $d(x,y)=d(y,x)\hspace{0.1cm}(\text{symmetry})\hspace{0.2cm} \forall x,y\in X\\$
  3. $d(x,y)=d(y,x)\hspace{0.1cm}(\text{symmetry})\hspace{0.2cm} \forall x,y\in X\\$

We can see that it verifies the second condition since $d\left ( x,y \right )=d\left ( y,x \right )=d^{*}\left ( x,y \right )=d^{*}\left ( y,x \right )$.

But how should I verify the first and the second one?

David C. Ullrich
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Martin N.
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1 Answers1

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The 3rd property should be the triangle inequality.

The proof for $d^\ast$ is trivial, because $d$ obeys all these properties, which are universal statements so they still hold when specialised to the subset $Y$.

Henno Brandsma
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  • When $Y\subseteqq X$, Y obeys all of the properties X, so does the triangle inequality, But how should I write it? – Martin N. Feb 16 '20 at 11:29
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    @MartinN. Take any $x,y,z \in Y$, then in particular $x,y,z \in X$ so $d^\ast(x,z) = d(x,z) \le d(x,y) + d(x,z) = d^\ast(x,y) + d^\ast(y,z)$, where the $\le$ comes from $d$ obeying the triangle inequality on $X$ and the rest is true by definition. – Henno Brandsma Feb 16 '20 at 11:32
  • Okay, thank you – Martin N. Feb 16 '20 at 11:35