Why is the topology on $\mathbb R$ always defined to be the set of "open" sets?

Question

The "standard topology" on $\mathbb R$ contains all the "open sets" on it. By open sets I don't mean "the sets that are in the topology", which would make the statement a tautology, but I mean the open intervals $(a,b)$, (which we can talk about prior to defining a topology).

My question is, why don't we define the "standard topology" as the topology generated by all the "closed intervals"? Why do we single out the open intervals only?

Answer 1

In addition to other answers let me write about a motivation behind the concept of topology.

Originally people were working with metric spaces $(X,d)$. But they quickly realized that there are properties sort of independent on $d$. For example for any $r\in\mathbb{R}$, $r>0$ consider

$$d_r:X\times X\to\mathbb{R}$$ $$d_r(x,y)=\min(d(x,y),r)$$

it can be easily seen that each $d_r$ is a metric and there are at least infinitely many different $d_r$ in the sense that there is an infinite subset $D\subseteq\mathbb{R}$ such that if $r,s\in D$, $r\neq s$ then $d_r(x,y)\neq d_s(x,y)$ for some $x,y\in X$.

But the following holds:

Lemma. Let $Y$ be a metric space and $f:X\to Y$ a function. Then $f$ is continuous in $d$ if and only if $f$ is continuous in $d_r$.

So even though these are different metrics they imply the same concept of continuity. That's because what really decides about continuity is those smaller and smaller values of $d$, not those "big" values of $d$. This idea led people to create the concept of topology. By isolating these properties of $d$ that decide about continuity.

And once it was created it was clear how to define a topology from a metric:

Definition. Let $(X,d)$ be a metric space. Call a subset $U\subseteq X$ "metric-open" if and only if for any $x\in U$ there is $r\in\mathbb{R}$ such that the open ball $B(x,r)=\{y\in X\ |\ d(x,y)<r\}$ is fully contained in $U$.

It can be shown that the set of all "metric-open" subsets of $X$ forms a topology on $X$. And this topology plays well with the metric:

Lemma. Let $X,Y$ be two metric spaces and $f:X\to Y$ a function. Then $f$ is coninuous in the metric sense if and only if $f$ is continuous in the topologies generated as defined above.

Note that it is essential to define the topology via open balls (corresponding to open intervals in $\mathbb{R}$). Otherwise (for example if you define via closed balls) the lemma is not true. Generally there's only one topology generated from closed balls (note that singletons are closed balls): the discrete topology, i.e. everything is open and everything is continuous. Boring.

Also note that topologies determine continuity uniquely in the following sense:

Lemma. Let $X$ be a set with two topologies $\tau_1, \tau_2$. The following are equivalent:
    1. $\tau_1=\tau_2$
    2. For any topological space $Y$ and any function $f:X\to Y$ we have that $f$ is continuous in $\tau_1$ if and only if $f$ is continuous in $\tau_2$.

So the concept of topology is the one that describes continuity uniquely. Unlike the concept of metric. I'm not saying that metrics are useless or worse. They are very important and have their applications elsewhere. But in this particular context topology is better.

With that in mind the definition of the standard topology on $\mathbb{R}$ comes from the standard metric on $\mathbb{R}$, i.e. the Euclidean one $d(x,y)=|x-y|$. Note that the definition you gave and the one I give are equivalent.

Answer 2

You can define any topology in terms closed sets. The reason why we usually think in terms of open sets is that local properties such as continuity is defined on an open interval.

Answer 3

Because the topology $\tau$ generated by the closed intervals is another topology, not the usual one. First of all, it contains the closed intervals, which do not belongs to the usual topology. Besides, every open interval $(a,b)$ belongs to $\tau$, since, if we take $N\in\mathbb{N} $ such that $a+\frac1N\leqslant b-\frac1N$,$$(a,b)=\bigcup_{n\geqslant N}\left[a+\frac1n,b-\frac1n\right].$$So, $\tau$ is a strictly larger set than the usual topology.

Actually, $\tau$ is just the discrete topology. This is trivial if yu include singletons in the set of closed intervals (after all, $\{a\}=[a,a]$). Even if you don't do that, you know that$$\{a\}=[a-1,a]\cap[a,a+1],$$and therefore $\{a\}\in\tau$. So, why would would bother to define $\tau$ as the topology generated by the closed intervals? It is much simpler to say that $\tau=\mathcal{P}(\mathbb{R})$.

Answer 4

Closed intervals can be singletons, so the topology generated by the closed intervals is the discrete topology. Not very interesting, not very useful.

Open intervals are the de facto standard of how we think about open sets. Every point lies "inside the set", and in $[0,1]$, neither $0$ nor $1$ lie inside the set from a topological perspective. They lie on the boundary of the set.

Finally, you can also argue that the topology is an order topology, and thus generated by open intervals by definition; or the metric-induced topology, which is generated by open balls which on the real numbers are the bounded open intervals.

Answer 5

By the axioms of a topology, if we'd use the proper closed intervals to generate a topology on $\mathbb{R}$, then for any $x$, $[x-1,x]$ would be "open" in this topology and so would $[x,x+1]$ be and so $[x-1,x] \cap [x,x+1] = \{x\}$ by the finite intersection axiom. So all singleton sets would be open as well, and thus all subsets of $\mathbb{R}$, as $A = \bigcup_{x \in A} \{\{x\}\}$ by the union axiom. So we would get the discrete topology, which is not what our intuition of $\mathbb{R}$ is, or continuity on it etc.

So we could take them as a generating set, but that would not give rise to an interesting topological space. The topology generated by the open intervals (and real analysis defined in terms of this topology) was an important example when defining a general concept of topological spaces; I'd say "open sets" are named after the open intervals (open as in not including bounding points, i.e. "open-ended") and it is pleasing that these terminologies coincide in this familiar domain.

Answer 6

Why don't we define the "standard topology" as the topology generated by all the "closed intervals"?

Because the "standard" topology should be the topology used as a "model" for the general definition. This "model topology" is the topology of the Euclidean spaces generated by the open balls (on the line, the topology generated by the open intervals).

The concept of topological space grew out of the study of the real line and euclidean space and the study of continuous functions on these spaces. (Munkres, p. 75)