Motivation. A few years ago we were using Armstrong's Basic Topology as a textbook for the topology course in my university, and right off the bat I had a huge conceptual problem regarding the two equivalent ways of defining a topology on a set declared above. I found the explanation given in the book confusing at the time since there were gaps in the arguments ("left to the reader") and the given arguments were a little bit too verbose for my taste (as opposed to notational, which some may deem excessive for this particular issue). In the lectures we have started from the second chapter, and as the neighborhood topology is introduced in the first chapter while the open set topology is introduced in the second, where the latter method was the only one we utilized, my concern became irrelevant, but it remained as a "mathematical ache", so to speak.
Now I am repeating the course to improve my GPA. So when I found the time I revisited this problem, and, surprise surprise, I believe I managed to write down a proof of the equivalence. I searched the web and math.SE for a proof (not necessarily very meticulously), but I could not find a full-fledged proof. Also Armstrong mainly refrains from using notation, which may result in someone studying the book to get confused by phrases such as "original open set" (p. 28). So I thought writing my proof here might prove useful for others too. Pointing out to mistakes in my proof, giving better proofs, or comparing (in terms of their purposes, specified at the end) these two ways of defining a topology are all appreciated. [Edit 01/04/23: This matter of comparison is really the content of the question. It seems this part got buried with the rest of what I wrote, hence I decided to reorder the sections.]
In what follows I will follows Armstrong quite closely in terms of definitions and arguments, though most of the notation is due to myself.
Comparison. In the book Armstrong compares the two topologies as follows (p. 27):
"The definition of a [$\mathcal{N}$-]topological space given in Chapter 1 fits quite well our intuitive idea of what a space ought to be . Unfortunately, it is not terribly convenient to work with, and our first job is to produce an equivalent, more managable, set of axioms."
It is immediate that the neighborhood topology emphasizes the nearness with respect to particular points more than the open set topology, as the former is really an assignment of neighborhoods to points, while the latter is a particular subset of the powerset. But I don't understand why the neighborhood topology is considered more intuitive than the open set topology. Also Armstrong claims that the open set topology is more practical. Could you explain how so, preferably with examples? Conversely could you point out to an instance where using the neighborhood topology is clearly more practical than using the open set topology?
[Edit 01/04/23] A better way of phrasing the question would be as follows: How do the intuitions that are associated to neighborhood topology and open set topology differ, and what are concrete examples when one intuition can be argued to be more beneficial than the other one?
It seems this question will not be reopened though I am still interested in different perspectives. It also seems to me that the common part of my question and the later question that got an answer is/was nothing more than the preamble of my question. For what it's worth, here is my tentative answer:
Here is one way to look at this matter: both neighborhoods and open sets are mathematical models of proximity. The intuition associated with neighborhoods is the aggregate of all those things that are close to an already predetermined thing. Thus I see a thing and take note of all the other things I see near it.
The intuition associated with open sets is a lump of things, without any one particular thing being the center of attention. Thus I look in a direction and realize there are some things that are close. Even the fact that the neighborhoods and open sets are equivalent has some associated intuition: One can look in a tray filled with candy; then focus on a certain part of the tray (open set) and then focus even further to a particular piece of candy (neighborhood); or else one can right away focus on a particular piece of candy (and see inevitably all the pieces of candy in the vicinity of said candy) and decide to relax his focus.
It seems to me Armstrong's statement that one is more intuitive and the other more useful is more of a personal statement than a statement of fact. Indeed one could say, matter of factly, that the frequency of usage of neighborhoods v open sets differ from discipline to discipline, which can be taken as symptomatic of which being more useful where. Here is a rough comparison of pros and cons of each model:
Neighborhoods come with basepoints; especially for "perturbation" statements this becomes useful (e.g. see https://math.stackexchange.com/a/3892256/169085).
Conversely open sets become more useful to work with when points are not available or not useful to consider (in particular this framework is more forgiving in terms of analogies, see e.g. https://math.stackexchange.com/a/1507071/169085; though it's also important to note that in borelesque categories having an analog of neighborhoods available (so called "standard measurable spaces") is quite powerful at keeping pathologies at bay).
Neighborhoods themselves need not be open; this at times reduces the things that need to be verified, at least at the syntactic level.
On the other hand one could also only consider open neighborhoods, which framework in certain cases inherits good traits from both frameworks. [End of Edit 01/04/23]
Definitions. First let us define the respective topologies (We signify a name related to the neighborhood construction by "$\mathcal{N}$-" and a name related to the open set construction by "$\mathcal{O}$-"):
- Let $X$ be a set, and let $\forall x\in X: \nu(x)(\neq\emptyset)\subseteq\mathcal{P}(X)$ be a collection of subsets of $X$ with the following four properties:
($\mathcal{N}_1$) $\forall N\in\nu(x):x\in N,$
($\mathcal{N}_2$) $\forall N,M\in\nu(x): N\cap M\in\nu(x),$
($\mathcal{N}_3$) $\forall N\in\nu(x): N\subseteq U\implies U\in\nu(x),$
($\mathcal{N}_4$) $\forall N\in\nu(x): N^\circ:=\{z\in N| N\in\nu(z)\}\in\nu(x)$ ($N^\circ$ is the ($\mathcal{N}$-)interior of $N$).
A member of $\nu(x)$ is a neighborhood of $x$; the assignment $\nu:X\to \mathcal{P}(\mathcal{P}(x)),x\mapsto\nu(x)$ is an $\mathcal{N}$-topology on $X$; and $(X,\nu)$ is an $\mathcal{N}$-topological space.
$O\subseteq X$ is $\mathcal{N}$-open if $\forall x\in O: O\in\nu(x) (\iff O=O^\circ$).
- Let $X$ be a set, and let $\tau(\neq\emptyset)\subseteq\mathcal{P}(X)$ be a collection of subsets of $X$ with the following three properties:
($\mathcal{O}_1$) $\emptyset,X\in\tau,$
($\mathcal{O}_2$) $\forall \{O_i\}_i\subseteq\tau:\bigcup_i O_i\in\tau,$
($\mathcal{O}_3$) $\forall\{O_k\}_{k=1}^N\subseteq\tau:\bigcap_k O_k\in\tau$.
A member of $\tau$ is an open set in $X$; $\tau$ is an $\mathcal{O}$-topology on $X$; and $(X,\tau)$ is a $\mathcal{O}$-topological space.
$\forall x\in X: N\subseteq X$ is an $\mathcal{O}$-neighborhood of $x$ if $\exists O\in\tau: x\in O\subseteq N$.
Outline. To show that these two ways of defining a topology on a set are indeed equivalent it suffices to show two things:
(1) An $\mathcal{N}$-topological space with the definition of an $\mathcal{N}$-open set gives an $\mathcal{O}$-topological space (and vice versa: an $\mathcal{O}$-topological space with the definition of an $\mathcal{O}$-neighborhood gives an $\mathcal{N}$-topological space):
$$[(\mathcal{N}_i)+\mathcal{N}\mbox{-open set}] \iff [(\mathcal{O}_j)+\mathcal{O}\mbox{-neighborhood}].$$
(2) When either construction is translated to the other one, open sets and neighborhoods are preserved, in the following sense:
("$\mathcal{ON}$-" is redundant) If we start with the neighborhood construction, then the $\mathcal{O}$-neighborhoods of the topology of $\mathcal{N}$-open sets, namely $\mathcal{ON}$-neighborhoods, are precisely the neighborhoods we started the construction with (and vice versa: ("$\mathcal{NO}$-" is redundant) if we start with the open set construction, then the $\mathcal{NO}$-open sets are precisely the open sets we started the construction with).
(TL;DR) Put differently, we have to show the following two statements:
- $\tau_\mathcal{N}:=\{O\subseteq X|O=O^\circ\}$ is the same $\mathcal{O}$-topology as $\tau$. Furthermore $\forall x\in X: \nu(x)=\nu_\mathcal{ON}(x)$, where $\nu_\mathcal{ON}(x)$ is the collection of $\mathcal{ON}$-neighborhoods of $x$.
- $\nu_\mathcal{O}$ is the same $\mathcal{N}$-topology as $\nu$, where $\forall x\in X:\nu_\mathcal{O}(x)$ is the collection of $\mathcal{O}$-neighborhoods of $x$. Moreover $\tau=\tau_{\mathcal{NO}}$.
Proof. First let us show (1).
($\implies$) As $\emptyset$ contains no elements it is $\mathcal{N}$-open vacuously. Any $x\in X$ has a nonempty collection of neighborhoods assigned to it by definition, so by ($\mathcal{N}_3$) is a neighborhood of any point. That takes care of ($\mathcal{O}_1$).
For ($\mathcal{O}_2$), if $x\in\bigcup_i O_i$, where $\{O_i\}_i$ is an arbitrary collection of $\mathcal{N}$-open sets, then for some $i$ we have $x\in O_i$. Since $\nu(x)\ni O_i\subseteq \bigcup_i O_i$, by ($\mathcal{N}_3$), $\bigcup_i O_i$ is a neighborhood of $x$.
($\mathcal{O}_3$) is immediately due to ($\mathcal{N}_2$) by induction: Take finitely many $\mathcal{N}$-open sets $\{O_k\}_{k=1}^N$ and $x\in\bigcap_k O_k$. Then $\forall k: O_k$ is a neighborhood of $x$, and consequently their intersection too is a neighborhood of $x$.
($\impliedby$) First observe that, as $\forall x\in X,\exists O(:=X)\in\tau:x\in O\subseteq X$, $X$ is an $\mathcal{O}$-neighborhood of any $x$, so that $\forall x\in X:\nu_\mathcal{O}(x)\neq\emptyset$. Fix $x\in X$.
For ($\mathcal{N}_1$) observe that if $N\in\nu_\mathcal{O}(x)$, then by definition $\exists O\in\tau: x\in O\subseteq N$. Next, if $N,M\in\nu_\mathcal{O}(x)$ we can find two open sets $O,P\in\tau$ with $N\supseteq O\ni x\in P\subseteq M$. Then we have $x\in O\cap P\subseteq N\cap M$, where $O\cap P$ is open, so that ($\mathcal{N}_2$) is satisfied. ($\mathcal{N}_3$) is immediate as well, since we can take the same open set $U$ with $x\in U\subseteq N$ for any larger set $N'\supseteq N$.
Finally let us see that the interior of an $\mathcal{O}$-neighborhood of $x$ is an $\mathcal{O}$-neighborhood of $x$ also. Let $N$ be an $\mathcal{O}$-neighborhood of $x$. Then $\exists O\in\tau: x\in O\subseteq N$. Let $z\in O$. Then $\exists U(:=O)\in\tau: z\in U\subseteq O$, so $O$ is an $\mathcal{O}$-neighborhood of $z$, which follows that $O\subseteq N^\circ$. Also $x\in N^\circ$, hence $N^\circ$ is an $\mathcal{O}$-neighborhood of $x$.
Next let us see why (2) holds.
("$\mathcal{ON}$-" is redundant) First we show that for any $x\in X$, the neighborhoods and the $\mathcal{ON}$-neighborhoods of $x$ coincide. Fix $x\in X$. If $N$ is a neighborhood of $x$, then, as $N^\circ=\{z\in N|n\in\mathcal{N}_z\}$, $N^\circ$ is $\mathcal{N}$-open. Since $N\in\nu_x$, by definition $x\in N^\circ$.
Observe that by definition a subset $U\subseteq X$ is an $\mathcal{ON}$-neighborhood of $x$ if $\exists$ $\mathcal{N}$-open $V$ such that $x\in V\subseteq U$. For $U:=N$, we can pick $V:=N^\circ$ by the previous paragraph.
Conversely, let $N$ be an $\mathcal{ON}$-neighborhood of $x$, i.e., $\exists$ $\mathcal{N}$-open $V$ such that $x\in V\subseteq N$. As $V$ is $\mathcal{N}$-open, it is a neighborhood of each of its points. In particular $V$ is a neighborhood of $x$. Then by ($\mathcal{N}_3$) $N$ too is a neighborhood of $x$.
("$\mathcal{NO}$-" is redundant) Let $O\in\tau$. Then $\forall x\in X\exists O_x(:=O)\in\tau: x\in O_x\in O$, so that $O$ is an $\mathcal{O}$-neighborhood of any of its points. This precisely means that $O$ is $\mathcal{NO}$-open.
Conversely, let $O$ be $\mathcal{NO}$-open. Then by definition $\forall x\in O: O$ is an $\mathcal{O}$-neighborhood of $x$. That is to say, $\forall x\in O,\exists O_x\in\tau:x\in O_x\subseteq O$. Then we have $\bigcup_x O_x=O$, and as each $O_x$ is open, by ($\mathcal{O}_2$), $O\in\tau$, which concludes the proof.
Continuity. This argument of course is extended to the definition of continuous functions:
A function $f:(X,\nu)\to(Y,\mu)$ is $\mathcal{N}$-continuous if $\forall x\in X,\forall M\in\mu(f(x)):f^{-1}(M)\in\nu(x)$. A function $f:(X,\tau)\to(Y,\sigma)$ is $\mathcal{O}$-continuous if $\forall V\in\sigma: f^{-1}(V)\in\tau$. Then we have the following:
$f:(X,\nu)\to(Y,\mu)$ is $\mathcal{N}$-continuous iff it is $\mathcal{O}$-continuous, provided that we pick $\tau:=\tau_{\mathcal{N}(\nu)}$ and $\sigma:=\tau_{\mathcal{N}(\mu)}$ according to their respective neighborhood topologies (observe that by the established equivalence above we need not check whether the same result holds when open set topologies are established initially, or when one set has a neighborhood topology and the other an open set topology).
Indeed, let $V\in\sigma$. Then $\forall y\in V,\exists V_y(:=V)\in\sigma:y\in V_y\subseteq V$, so that $V$ is a ($\mathcal{O}$-)neighborhood of each of its points. If $f$ is $\mathcal{N}$-continuous, then $f^{-1}(V)$ is a ($\mathcal{O}$-)neighborhood of each of its points. This follows that $\forall x\in f^{-1}(V),\exists U_x\in\tau:x\in U_x\subseteq f^{-1}(V)$, which in turn means that $f^{-1}(V)=\bigcup_{x\in f^{-1}(V)}U_x\in\tau$.
Conversely, fix $x\in X$. If $M$ is a neighborhood of $f(x)$, consider $M^\circ$. $f(x)\in M^\circ$ and $M^\circ$ is ($\mathcal{N}$-)open by definition. If $f$ is $\mathcal{O}$-continuous, then $f^{-1}(M^\circ)\in\tau$. Since $f(x)\in M^\circ\subseteq M$, we also have $x\in f^{-1}(M^\circ)\subseteq f^{-1}(M)$. As $f^{-1}(M^\circ)$ is ($\mathcal{N}$-)open, it is a neighborhood of each of its points. In particular $f^{-1}(M^\circ)$ is a neighborhood of x. Then by ($\mathcal{N}_3$) $f^{-1}(M)$ is a neighborhood of $x$ as well. Hence continuous functions too are preserved.
