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This is the definition of neighborhood in the book Basic Topology by M.A. Armstrong. I am wondering whether $\{x\}$ is a neighborhood of $x$ according to the definition above. As it follows "Let $X$ be a topological space and call a subset $O$ of $X$ open if it is a neighborhood of each of its points.", is there any contradiction?

Arctic Char
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Hanson Lu
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  • This is actually the definition of open set. I guess you should have defined first what neighbourhoods are – Lorenzo Pompili Jul 01 '21 at 15:43
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    In any case, it is obvious that ${x}$ in general is not always a neighbourhood of $x$ in a topological space (nor in a metric or normed space...) – Lorenzo Pompili Jul 01 '21 at 15:45
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    It is only a characterization of open sets. When you say 'let X be a topological space', you already know the open sets, they are the elements of the given topology on $X$. Now a set $N$ is a neighborhood of $x$ if there is an open set $U$ such that $x\in U\subset N$. It follows that a set is open iff it is a neighborhood of each of its points. – Gribouillis Jul 01 '21 at 15:48
  • Singleton sets are only open in the Discrete Topology/metric spac,e which is a pretty boring place – Alan Jul 01 '21 at 15:51
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    @Gribouillis forgive me, I thought you were the one that posted the answer, but allow me to disagree. In the book that he mentions, the author first gives the definition of neighbourhoods (axiom 1.3). Then it defines what are the open subsets. This is highly non-standard in my opinion and it is better to define open sets first, but it works as well as the way that you are saying, that is what every mathematician studies – Lorenzo Pompili Jul 01 '21 at 16:12
  • @LorenzoPompili I haven't read the book but it looks indeed non standard to define "open set" after "topological space". – Gribouillis Jul 01 '21 at 16:18
  • @Gribouillis He in fact defines "topological structure" the family of all neighborhoods (that are sets that satisfy certain axioms). Then he says that a topological space is a set equipped with a topological structure, and then he defines open sets. Again, I agree with you, this is non standard and maybe it is suitable for some points of view, but it seems it works – Lorenzo Pompili Jul 01 '21 at 16:23
  • You should write down Armstrong's definition. – Paul Frost Jul 02 '21 at 11:01
  • This is actually the first time "neighborhood" ever appears in that book, which is really puzzling. I might as well accept that this is a non-standard way. Thank you all. – Hanson Lu Jul 02 '21 at 12:10

2 Answers2

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Edit: this answer refers to an old version of the post.

To sum up, the one that you wrote is not the definition of neighborhoods, but that of open sets (see the comments below your question).

If you are referring to the definition (1.3) of neighborhoods in the book, then you may check that for instance, the family of neighborhoods given by the Euclidean topology of $\mathbb R$ satisfies the definition of neighborhoods, but clearly, $\{x\}$ is never an open subset of $\mathbb R$.

Let me say that in general it could be that $\{x\}$ is open - thus it is a neighbourhood of $x$ - maybe just for some x of the topological space, but this is not what happens in general. About $\{x\}$, you can very often say that it is closed, for example if the space $X$ is a metric space

Lorenzo Pompili
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From the comments I gather ( I don't have the book) that your text starts by axiomatically defining a so-called "neighbourhood space" (as defined in this question e.g. or on Wikipedia here). This is perfectly valid (but somewhat old-fashioned). So for every $x \in X$ we have a family $\mathcal{N}(x)$ of subsets of $X$ (all containing $x$) that forms a filter and such that all $\mathcal{N}(x), x \in X$ together have some coherence condition.

As the families $\mathcal{N}(x)$ are given it makes sense to define $O$ open iff

$$\forall x \in O: O \in \mathcal{N}(x)\tag{1}$$

which is what the phrase "$O$ is a neighbourhood of each of its points" translates to, mathematically.

And $\{x\}$ is thus open iff for that $x$ we happen to have that $\{x\} \in \mathcal{N}(x)$, which indeed can be the case (it's not ruled out by the axioms and it does imply that any set $N$ that contains $x$ is also in $\mathcal{N}(x)$ as well) but this does not happen in the standard Euclidean system of neighbourhoods in $\Bbb R$ or $\Bbb Q$, say. It will depend on the neighbourhood system you're given or that has been defined.

Such $x$ where it does happen are called isolated points of $X$, and they're topologically not very interesting. If all $x \in X$ are isolated points, $X$ is discrete, and all subsets of $X$ are open, as $(1)$ then trivially implies.

Henno Brandsma
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