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I know that a metric is a way of measuring distance that can be generalized to sets outside of the real numbers. I also know that the space is simply the set you are working in. So is a metric space just the outcome of putting your set through the metric?

For example, if we have $(\mathbb{R}, d_0)$ with $d_0$ being the discrete metric does $\mathbb{R}$ simply become $\left\{0, 1\right\}$ because they are the only possible outcomes, like taking an open or closed ball? More likely, a set is the same no matter what metric it is under, but then why can the same set be open in one metric and closed in another? (asides from induced metrics I kind of understand that)

I've done quite a lot of work with metrics at this point and still feel uneasy with them, especially with non-Euclidean metrics since the Euclidean metrics are the ones I've worked with the most. I guess my question is what do spaces look like once they are under the effect of different metrics?

I'm sorry this is a badly formatted and vague question, but I don't really know what I'm trying to ask. I just can't get my head around the same space under different metrics and what they look like. If anyone has examples or explanation that might give me a new understanding I would be very grateful. If this question should be deleted, tell me and I will.

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    $\begingroup$ $\mathbb{R}$ with the discrete metric is still just $\mathbb{R}$, but that doesn't help us figure out what it looks like. To actually picture it, imagine you are at some point $x \in \mathbb{R}$. Then $d(x,x)=0$, and $d(x,y)=1$ for every $y \neq x$. So if you look around you, every other point looks like it's distance $1$ from you. This is true at every point. It's really hard to picture this, because it's not Euclidean. Non-Euclidean metrics on $\mathbb{R}^n$ distort the space from what we are used to. $\endgroup$ – kccu May 9 '17 at 18:46
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    $\begingroup$ This question is probably too broad, but to answer your question about the discrete metric specifically: I like to think of $(\mathbb{R},d_0)$ as an infinite number of "separated" or "spread out" points, as many as there are real numbers. (Think of a cloud of little particles, with some distance separating each.) This is in part useful because it shows how each point is its own neighborhood, although of course it isn't faithful because distances between different points appear to vary even though they should all be equal to $1$. $\endgroup$ – Alex Provost May 9 '17 at 18:48
  • $\begingroup$ Thanks for the responses. The idea of considering a generic point in the set helps a lot especially in combination with Alex's visualisation of the discrete metric as separated dots. $\endgroup$ – Mike A May 9 '17 at 19:05
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    $\begingroup$ "I guess my question is what do spaces look like once they are under the effect of different metrics?" Well, look, e.g., at en.wikipedia.org/wiki/Ultrametric_space $\endgroup$ – Michael Hoppe May 9 '17 at 19:22
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    $\begingroup$ @MikeA You're definitely welcome. $\endgroup$ – Michael Hoppe May 9 '17 at 19:44
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A metric space is a formal object. It doesn't (and doesn't need to) "look like" anything. Trying to picture them can be useful, but most of times is of little importance. What is important is why you are considering a given metric. For example, if you want to measure how far away a real function on the interval is from the $0$ function, in some applications you want to consider that it is far if there is a value of $f$ which is far away. In others, you want to consider that it is is not very far if there is a small area underneath its graph.

Since it is a formal object, it is psychologically natural to try to grasp the concepts in terms of what you already understand. But that completely misses the point of the power of the abstraction, which is to surpass the confort zone of what you already understand.

That being said, I think that there is an implicit mathematical question in your soft question. Maybe what would make you happy is among the following lines

Given a metric space $(X,d)$, is there a way to visualize it inside $\mathbb{R}^3$?

Which can be satisfactorily restated mathematically as

Given a metric space $(X,d)$, is there an isometry of $X$ with a subspace of $(\mathbb{R}^3,d_{\text{euclid}})$?

An isometry, if you are not familiar with the term, is a function which preserves distance.

For example, the abstract metric space $X=\{A,B,\text{Lettuce}\}$ with discrete metric admits an isometry to a subset $T$ of $\mathbb{R}^3$: take the vertices of an equilateral triangle in the plane.

However, not every metric space can be visualized like that. This is of course trivial if you consider a set of very big cardinality and put any metric you want on it (e.g., the discrete). However, even metrics on subsets of $\mathbb{R}^3$ itself may not be realized as a subspace of $\mathbb{R}^3$. For example, the same discrete metric in $\mathbb{R}$ cannot be visualized in $\mathbb{R}^3$-*.

Therefore, you can't reasonably imagine $\mathbb{R}$ with the discrete metric as a cluster of points in space in the way I mentioned. But that is not so harsh as it sounds: you don't need to.

*For a proof, one way to see that is by noting that a subspace of a secound-countable topological space needs to be secound-countable. But the discrete metric on an uncountable set is not secound countable. This shows further that not even topologically it can be realized as a subset of Euclidean space.

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  • $\begingroup$ Thank you for thoughtful response. Some of it flew over my head a bit (countability) but I definitely see your point on how it isn't always appropriate. $\endgroup$ – Mike A May 9 '17 at 20:08
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Different metric on the same set sometimes induces different topology, you can think a metric on a set is like some sort of partition of the set into open sets.

You may think that different metric on the same set partitions the set differently.

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  • $\begingroup$ Since the OP seems to lack a general understanding of metric spaces and your answer involves even topologies which are more advanced than metrics, I downvoted your answer. It doesn't seem to be helpful, at least if you don't try to explain a little more $\endgroup$ – Jakob Elias May 9 '17 at 18:54
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    $\begingroup$ It's fine, your wish, i think i made myself clear, even if he does not understand topology he would understand what i am trying to say. $\endgroup$ – Arpan1729 May 9 '17 at 18:56
  • $\begingroup$ I appreciate the response but Jakob is right in that I have not dealt with topology in general during my course but only metric spaces. I understand what you mean to some degree though. Thanks for the response. $\endgroup$ – Mike A May 9 '17 at 19:02

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