# Do all subsets of metric spaces have boundry points?

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.

• $\exists(X,d)$? Also, your definition of a boundary point is wrong, you're claiming that $A^c$ intersects $A$. – Jack M Feb 12 '14 at 19:14
• @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 – Mikkel Rev Feb 12 '14 at 19:33
• 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". – Jack M Feb 12 '14 at 19:36