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I don't really understand why the topology on the metric space has to be defined in terms of open balls when closed balls seems to work just fine.

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    @Arthur: But the closed balls don't "generate" the closed sets, in the same way that open balls generate open sets. – 2017-01-12
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    What's your definition that "works just fine"? We can't comment if you don't show us what you are proposing (although several people have $\ddot{\smile}$). – 2017-01-12
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    Do you take the closed ball of radius $0$ to be open? Because then *every* set is open, being a union of the singletons contained in it, which were taken to be open. – 2017-01-12

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You don’t in general get a base for a topology. In $\Bbb R$, for example, with the usual metric, the closed balls of radius $1$ centred at $0$ and at $2$ are the intervals $[-1,1]$ and $[1,3]$, respectively. Their intersection is the singleton set $\{1\}$, which does not contain a closed ball around $1$.

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    What's wrong if the singleton set is declared open? – 2017-01-12
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    @user2277550: That’s not relevant to your original question. I just showed that the family of closed balls doesn’t work just fine, because it isn’t necessarily a base for *any* topology, let alone the one generated by the metric. If you do declare all singletons open, you get the discrete topology; it’s metrizable, but not by the original metric. – 2017-01-12
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    I'm sorry [a-$r_1$, a+$r_2$] seems to form a basis for $a,r_1,r_2 \in \mathbb{R}$ – 2017-01-12
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    @user2277550: If you allow $r_1=r_2=0$, you get a base for the discrete topology. However, this has nothing to do with your original question, The closed balls with respect to the usual topology are the sets $[a-r,a+r]$ with $a,r\in\Bbb R$ **and** $r>0$, and I just showed you that they are not a base for any topology. – 2017-01-12
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Let $\tau_{_X}$ be the collection of open sets on a topological space $X$. By definition, a basis for the topology on $X$ is a subset $B \subset \tau_{_X}$ such that any element of $\tau_{_X}$ can be realized as a union of some collection of elements from $B$. The problem in this case is that closed balls aren't open sets in the metric topology, so the collection of closed balls cannot be a base. If you try to treat closed balls as one would open balls and expect the usual stuff to be true, you're out of luck. For example, in $\mathbb{R}$, notice $\displaystyle \bigcup_{n=2}^\infty [-1 \! + \! 1/n, \ 1 \! - \! 1/n] = (-1, 1)$, which isn't closed.

However, it's worth noting that we can redefine a topological space w.r.t. closed sets. In particular, a topological space $X$ can be defined as a set $X$ together with a collection $\sigma_{_X} \subset \mathcal{P}(X)$, which we'll call the "closed sets" such that:

  • $\emptyset, X \in \sigma_{_X}$
  • Any union of finitely many elements of $\sigma_{_X}$ is an element of $\sigma_{_X}$.
  • Any arbitrary intersection of elements of $\sigma_{_X}$ is an element of $\sigma_{_X}$.

Recall there is a bijective correspondence between closed sets and open sets: if $U \subset X$ is open, then $X \setminus U$ is closed. It is from this that the "new" definition for a topological space arises.

Even in this context, the base you want still wouldn't quite work. The corresponding base would be the collection $\{ X \setminus U \ | \ U \text{ is open } \}$, and one can show that any other closed set can be written as an intersection of such sets.

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    How can an unbounded closed subset of $\Bbb{R}^n$ (like one of the coordinate axes) be an intersection of closed balls? – 2017-01-12
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    @RobArthan, oh, you're right. The base wouldn't be closed balls. It'd be the whole space with an open ball removed. Darn. – 2017-01-12
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    Never mind, the OP (who has failed to clarify the question) has given you the green tick anyway. – 2017-01-12
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    Fixed nevertheless. – 2017-01-12
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    @KajHansen again why shouldn't (0,1) be closed? Let it be closed. So do we have a new topology now? – 2017-01-12
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    It cannot be closed if we're talking about the topology induced by the Euclidean metric @user2277550. A closed set has all of its limit points, and one can show that $0$ and $1$ are limit points. There are certainly *other* topologies that we can put on $\mathbb{R}$ so that $(0,1)$ can be a closed set. For example, the discrete topology on a set $X$ is such that every subset $U \subset X$ is open. Hence, every subset $U \subset X$ is closed as well. So $(0, 1)$ would be closed under the discrete topology on $\mathbb{R}$. There are surely others. – 2017-01-12
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Think about the natural topology on the plane. If $x$ is a boundary point of a closed ball $B$, there is no any closed ball centered at $x$, which is a subset of $B$. Then closed balls can not form any basis of neighbourhoods on a plane.

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You can define a topology in terms of what its closed sets are, just as well as the open ones. It's just that we usually don't, because open sets are nicer most of the time. In the specific setting of defining the topology of a metric space using closed balls as a (sub)basis, the fact that arbitrary intersections of closed sets are closed, but only finite unions, means that it is extra cumbersome because we need to take care of the big closed subsets of our space, like the entire space itself, or any infinite set of isolated points (without any limit points). Those are not always a closed ball, or a finite union of such, so the definition becomes longer and probably less wieldy.

One case where we do usually define the topology of a space in terms of its closed sets is the Zarisky topology.