Say, for example, let $(X, d)$ be a metric space, then the basis of the topology which the metric d induced is open balls $\mathbb{B} = \{\mathbb{B}_d(x, r), x\in X, r>0)$.

Then those open balls cover the whole $X$.

Since there exists a ball $\mathbb{B}(x, 1)$, such that $x\in \mathbb{B}(x, 1)$, for every $x\in X$.

But how could we know, every such open ball $\mathbb{B}(x, 1)$ is actually in $X$? Can it possible that some such open balls are only partial in $X$? Or what open in open ball here just means open in the subspace $X$?

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    $\begingroup$ By definition of open ball $\mathbb{B}_d(x,r):=\{y\in X\text{ such that }||y-x||_d<r\}$ it's a subset of $X$ so it is properly contained in it. $\endgroup$ – Sebastian Cor Nov 21 '19 at 3:08
  • $\begingroup$ @SebastianCor But for example let [0, 1) be a metric space induced by R usual metric on [0, 1), then every ball containing 0 is not properly contained in [0, 1) $\endgroup$ – Cathy Nov 21 '19 at 3:24
  • $\begingroup$ Considering $[0,1)$ as its own space, every ball containing $0$ is properly contained in $[0,1)$. For example, $\mathbb B(0,1) = \{x\in[0,1):|x|<1\} = [0,1)$. $\endgroup$ – Math1000 Nov 21 '19 at 3:44
  • $\begingroup$ @Math1000 So the B(0,1) is open somehow means open in the subsapce [0, 1), not the whole space, say such as R? $\endgroup$ – Cathy Nov 21 '19 at 3:48
  • $\begingroup$ Re: Last comment above. Exactly. There is no "whole space" in the sense that $any$ metric space is a proper subspace of a larger metric space. $\endgroup$ – DanielWainfleet Nov 21 '19 at 3:53

By definition, an open ball in a metric space is given by

$$ \mathbb{B}_d(x,r) \;\; =\;\; \{p \in X \; | \; d(x,p) < r\} $$

hence by definition we have that $\mathbb{B}_d(x,r) \subseteq X$. Now, if you are in some metric space $X$, you may be interested in working in some subspace $Y$ of $X$, in which open balls in $Y$ will be those induced by $X$. In other words, we will have

$$ \mathbb{B}_d^Y(x,r) \;\; =\;\; \{p \in Y \; | \; d(p,x) < r\} \;\; =\;\; \mathbb{B}_d^X(x,r) \cap Y. $$

It's therefore important that when you talk about open balls that you are clear about whether they are open balls with respect to the total metric space $(X,d)$ of interest, or if you're referring to a subspace $Y$ of $X$.


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