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Consider the metric space $(X,d)$. I feel like $d(d(x,y),0) = |d(x,y)-0| = |d(x,y)| = d(x,y)$. The last step (removal of the |.|) follows due to $d(x,y) \geq 0$. For the rest of the proof, it almost seems like an exploitation of the notation and not something that follows from the axioms. I am considering a a general metric space, so that does not seem very convincing.

oldsailorpopoye
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    $d(d(x,y),0)$ is, in general, not defined. $d$ wants its arguments to be elements of $X$, while $d(x,y)$ is a real number. Unless $X \subset \mathbb R$, you cannot put real numbers in place of $x$ or $y$ in $d(x,y)$. – lisyarus May 28 '20 at 16:05
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    Your first equality $d((d(x, y), 0) = |d(x, y) - 0|$ doesn't really makes sense since you never mentioned an explicit form of $d$. If you are taking $X = \mathbb{R}$ and endowing it with the usual metric, then yes the first equality is true. But then the last equality becomes $|x - y|$ i.e. $d(d(x,y),0) = |x - y|$. – saru May 28 '20 at 16:08
  • @lisyarus Right, and then we can have the argument follow what I just did above. I didn't take into account the fact that $d : X \times X \to \mathbb{R}$ – oldsailorpopoye May 28 '20 at 16:08
  • Suppose that $X$ is the set of vertices of a connected undirected graph $G(X,E)$ and $d(x,y)$ is the length of the shortest path between $x$ and $y$ in $G$. It's easy to check that $(X, d)$ is a metric space. Now let $x,y \in X$ and consider the following question: what is $d(d(x,y),0)$? – Rick May 28 '20 at 16:11
  • The statement is true for general metric spaces, but you should be careful with notation. The outer $d$ on the left hand side is on the space $\mathbb{R}$ while the other $d$s are the metric on $X$. It is more precise to write $d_E(d(x,y), 0) = d(x,y)$, where $d_E$ is the euclidean metric on $\mathbb{R}.$ – Jens Renders May 28 '20 at 16:25

1 Answers1

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N.B. I am assuming that you meant just one metric $d$ on $X$ here and not two different metrics.

In a general metric space, what you wrote doesn't make complete sense. If $X$ is not a subset of $\mathbb{R}$ or $\mathbb{C}$, then what is $0$? A metric $d$ on $X$, before it satisfies the metric properties, is first a function $d : X \times X \to \mathbb{R}$ so it has to "feed" on elements of $X$. If $0 \notin X$, then $d(\cdot,0)$ doesn't really make sense as a metric.

On the metric part itself, just like in the comments, if you are considering $X = \mathbb{R}$ and the usual metric on $\mathbb{R}$, then yes it does makes sense to have

$$d((d(x, y), 0) = |d(x, y) - 0| = |d(x,y)| = d(x, y).$$

In fact, if it is the usual metric, $d(x,y) = |x-y|$ and so $d((d(x, y), 0) = | x- y|$.

saru
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  • I disagree with how you put it. It should just be a small remark that OP is using 2 metrics and calling them both $d$. If he wrote $d_E(d(x,y), 0) = d(x,y)$ then it would make sense and it is also true. There is no problem with $0$ as he is comparing $0$ with the output of $d$, both are always in $\mathbb{R}$. – Jens Renders May 28 '20 at 16:21
  • I don't think the OP ever mentioned that there are two metrics here. This is reflected in the OP's first sentence, "consider the metric space $(X, d)$". Furthermore, if your definition of $d_E$ is the usual metric, this is precisely the second part of my answer. – saru May 28 '20 at 16:30
  • This is not precisely the second part of your answer. In your answer you go the the case $X = \mathbb{R}$ and $d = d_E$ (yes I write $d_E$ for euclidean metric on $\mathbb{R}$). That is a very specific case. It is still true in general that $d_E(d(x,y),0) = d(x,y)$. – Jens Renders May 28 '20 at 16:33
  • OP indeed does not seem to realize that he is using 2 metrics (on 2 spaces), so just pointing that out is all that is needed IMO. – Jens Renders May 28 '20 at 16:34
  • Well, I had to go to $X = \mathbb{R}$ because I'm assuming the OP meant we are dealing with only one metric $d$ here and not two metrics. If we are dealing with two metrics $d_E : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ and $d: X \times X \to\ \mathbb{R}$ and defining as you put it, then yes it is OK regardless what $X$ is. – saru May 28 '20 at 16:35
  • I take this viewpoint because people often denote different metrics with the same symbol $d$, just like they denote different sums or multiplications (or powers) with the same symbol ($+$, $\cdot$). This is sloppy though. – Jens Renders May 28 '20 at 16:37
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    I agree with you on this one. I have edited my answer and put a disclaimer above. – saru May 28 '20 at 16:37
  • @JensRenders Right, I did not realize that I mixed up two different metrics. $d$ is an arbitrary metric and not euclidean, in particular.

    All of it made sense when user lisaryus mentioned in comment #1 under the post (/question). Please look into comment #3 (where I wrote that it all was clear to me)

     – oldsailorpopoye Jun 07 '20 at 22:37
  • @JensRenders Thank you for the answer. Since it does not solve the question, I am not accepting it. I just mixed up the metrics. – oldsailorpopoye Jun 07 '20 at 22:43