On different definitions of neighbourhood.

I am going through the basics of topology, mainly to refresh them. I had taken a course some years ago but never used topology actively. So I am reading Munkres's Topology. I have noticed that he defines the notion of a neighbourhood in a way different to what I remember to have seen before: he defines a neighbourhood $\mathcal{N}$ of a point $x$ of a topological space $X$ to be an $\bf{open}$ set containing $x$, while I remember that neighbourhoods are otherwhere defined as sets $\bf{containing}$ an open set that contains $x$.

My question is, how much does one lose (or gain) from using Munkres's definition over the other one? Are there significant changes in the theory, that I should consider before continuing reading this text? Thanks.

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One loses a lot of convenience by demanding a neighbourhood be open, but nothing in the theory. Instead of saying "neighbourhood", you need to say "set containing a neighbourhood" or something along those lines in a lot of places. The family of neighbourhoods of a point no longer is a filter in general (and that's damn inconvenient, yuck$^{15}$). – Daniel Fischer Jul 20 '14 at 15:27
Whichever definition you use, you'll find that you need to refer to the other notion occasionally. I like to use the convention that neighborhoods need not be open; that way, when I want the other notion, I just say "open neighborhood". If I required neighborhoods to be open, then the other notion would be "set containing a neighborhood" (as Daniel Fisher said). I like the relative brevity of "open neighborhood" (and I agree with Daniel about liking filters). – Andreas Blass Jul 20 '14 at 15:37
I stumbled over this distinction a long time ago. Now I try to explicitly write 'open neighbourhood'. – copper.hat Jul 20 '14 at 15:38

Let's call the two definitions "Munkres neighborhood" and "broader neighborhood."

Let $\tau$ be a topology on $X$ and $x\in X$. Define a new topology on $X$ as $$\tau_x=\left\{V\subseteq X\mid x\notin V \lor \left(\exists U\in\tau:\, x\in U\subseteq V\right)\right\}$$ This is easily proved to be a topology.

So the Munkres neighborhood of $x$ in $(X,\tau_x)$ coincides with the definition of the broader neighborhood of $x$ in $(X,\tau)$.

$\tau_x$ is a localized version of the topology $\tau$ at $x$, and we have that $\tau = \bigcap_{x\in X} \tau_x$. This means $f:(X,\tau)\to Y$ is continuous if and only if for all $x\in X$, $f:(X,\tau_x)\to Y$ is continuous.

The topology $\tau_x$ lets you define "continuity at $x$." That is, in point-set topology, we can only define a function as continuous, not "continuous at a point." In metric spaces, though, we can define continuity at a point $x$, and it coincides with the definition $f:(X,\tau_x)\to Y$. So if we use $(X,\tau_x)\to Y$ being continuous to define "continuity at a point," we see that if $f$ is continuous at every point, it is continuous.

I suspect Munkres' definition is the most common in modern works. Perhaps in early topology days, people were concerned about generalizing the idea of continuity at individual points. The above construction shows that "continuity at a point" is, perversely, just a special case of the more global notion of continuity. Rather than localizing the definition of "continuity" we localize the topology. Munkres definition still supports the broader definition by moving to this localized topology $\tau_x$.

Note, also, that you can get a dual notion, $$\tau^x=\{U\in \tau\mid x\in U\lor U=\emptyset\}$$

Then $\tau = \bigcup_{x\in X} \tau^x$. (Usually, the union of topologies is not a topology, but in this case, the union can be seen to be the original topology. More generally, be careful to take joins of topologies, however.)

$\tau^x$ is the topology containing the Munkres neighborhoods of $x$ (and the empty set) and is related to continuity of functions to $x$, a concept which is foreign to people familiar with the metric definition of continuity. What would it mean for $f^{-1}(U)$ to be open only for the open sets containing $x$?

However, $\tau^x$ has the same property: If $f:Y\to X$ is continuous to $x$ for each $x\in X$, then $f$ is continuous.

If $(X,\tau)$ and $(Y,\rho)$ are two topologies, then $f:(X,\tau)\to (Y,\rho)$ is continuous if and only if, $f:(X,\tau_x)\to(Y,\rho^y)$ is continuous for all $x\in X,y\in Y$.

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