I am learning about metric spaces.

I failed to find an aunambiguous answer to my question on Google. So this is the right place to ask:

Assume (X,d) is a metric space, and $A \subset X$. If I understand correctly, as long as $(X,d)$ is a metric space $\implies X \neq \varnothing$? If that is the case, then the following may or may not be true:

«$A$ is a strict subset of $X$ $\implies$ $A$ has points $x \in X : \forall \varepsilon > 0, \,\exists B(x,\varepsilon) : B(x,\varepsilon)\cap A \neq \varnothing \wedge B(x,\varepsilon)\cap A^c \neq \varnothing$»

If you know, will you kindly tell me if this is true, and why that is.

Thank you for your time.

Kind regards,


  • $\begingroup$ $\exists(X,d)$? Also, your definition of a boundary point is wrong, you're claiming that $A^c$ intersects $A$. $\endgroup$ – Jack M Feb 12 '14 at 19:14
  • $\begingroup$ Ac∩A=∅ hold for any A! So this does not have sense. $\endgroup$ – Emin Feb 12 '14 at 19:16
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    $\begingroup$ Also, a simple counterexample to this is to take the discrete metric, with which at any point in any set, there are balls which intersect no other points at all, and thus in particular do not intersect both the set and its complement. I can't think of any less boring counterexamples off the top of my head. $\endgroup$ – Jack M Feb 12 '14 at 19:21
  • $\begingroup$ @JackM, thank you for pointing out the error about the intersection. I believe I fixed that now. Furthermore, I thought $\exists (X,d) \iff$ «there exists a $(X,d)$», which embedded in the rest of the sentence would translate to: «Assume there exists a set X, and a metric d, a metric space, and that $a \subset X$». Is that incorrect? If so I would very much like to know what is wrong, so that I can correct my notation. Thank you for your time.M $\endgroup$ – Mikkel Rev Feb 12 '14 at 19:33
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    $\begingroup$ It's not really what the symbol $\exists$ was intended for. You use it to make a claim that something exists whose existence might not be obvious, for example, $(\exists x\in A\forall y\in A) x\geq y$ asserts that $A$ contains a maximum, which isn't necessarily true. You simply wish to assign a name to a metric space that you're introducing, you're not making the assertion "there exists at least one metric space". $\endgroup$ – Jack M Feb 12 '14 at 19:36

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