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I am still unsure as to how to go about proving if something is a metric space or if a specified distance function defines a metric space. I am attempting to tackle the following and would like any tips\corrections if possible. I know that a distance function must satisfy the following:

  1. $d(x,y)\geq 0$ (equals $0$ if $x=y$)
  2. $d(x,y)=d(y,x)$
  3. $d(x,z) \leq d(x,y) + d(y,z)$

Suppose $d: X \times X \to \mathbb{R}$ is a distance function. Are the following also distance function $\rho: X \times X \to R$:

  1. $\rho(x,y)= (d(x,y))^2$
  2. $\rho(x,y)= (d(x,y))^{1/2}$
  3. $\rho(x,y)= 3d(x,y)$
  4. $\rho(x,y)= (d(x,y))^{1/2} + 2d(x,y)$

Proof:

  1. Clearly the first two properties hold. The triangle inequality comes down to showing if $d(x,z)^2 \leq d(x,y)^2 + d(y,z)^2$. We run into a contradiction if we let $X$ be the real line and consider the points $x=1, z=-1, y=0$.
  2. Not quite sure
  3. Clearly, the first two properties hold. When we face the triangle inequality, we can factor out the $3$ and our inequality is our usual triangle inequality which we assume holds since we assume $d$ is a distance function.
  4. Because of $2$ I am unsure about it. I believe it will\will not hold depending on this problem.
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For 1., let $X$ be the real line with the usual metric, and take $x=-1$, $z=1$, and $y=0$.

For 2., $$\rho(x,z) \le \sqrt{d(x,y)+d(y,z)} \le \sqrt{d(x,y)} + \sqrt{d(y,z)} = \rho(x,y) + \rho(y,z).$$

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