The following axioms of a Topological space is from Wikipedia:

Neighbourhoods definition

This axiomatization is due to Felix Hausdorff. Let $X$ be a set; the elements of $X$ are usually called points, though they can be any mathematical object. We allow $X$ to be empty. Let $N$ be a function assigning to each $x$ (point) in $X$ a non-empty collection $N(x)$ of subsets of $X$. The elements of $N(x)$ will be called neighbourhoods of $x$ with respect to $N$ (or, simply, neighbourhoods of $x$). The function $N$ is called a neighbourhood topology if the axioms below are satisfied; and then $X$ with $N$ is called a topological space.

  1. If $N$ is a neighbourhood of x (i.e., $N \in N(x)$), then $x \in N$. In other words, each point belongs to every one of its neighbourhoods.
  2. If $N$ is a subset of $X$ and contains a neighbourhood of $x$, then $N$ is a neighbourhood of $x$. I.e., every superset of a neighbourhood of a point $x$ in $X$ is again a neighbourhood of $x$.
  3. The intersection of two neighbourhoods of $x$ is a neighbourhood of $x$.
  4. Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$.

Then there is an example given:


  1. $X = \{1, 2, 3, 4\}$ and collection $\tau = \{\{\}, \{1, 2, 3, 4\}\}$ of only the two subsets of $X$ required by the axioms form a topology, the trivial topology (indiscrete topology).
  2. $X = \{1, 2, 3, 4\}$ and collection $\tau = \{\{\}, \{2\}, \{1, 2\}, \{2, 3\}, \{1, 2, 3\}, \{1, 2, 3, 4\}\}$ of six subsets of $X$ form another topology.
  3. $X = \{1, 2, 3, 4\}$ and collection $\tau = P(X)$ (the power set of $X$) form a third topology, the discrete topology.
  4. $X = \mathbb{Z}$, the set of integers, and collection $\tau$ equal to all finite subsets of the integers plus $\mathbb{Z}$ itself is not a topology, because (for example) the union of all finite sets not containing zero is infinite but is not all of $\mathbb{Z}$, and so is not in $\tau$.

Here is my question: From axiom 1, it is said that an element $x$ in the set must belong to any of its neighborhoods, but in examples 1 and 2, the empty set is a neighboorhood. I don't understand it. I am not a math student, I'm just curious.

  • 4
    $\begingroup$ Those examples are from the "open sets definition" section of the page. The sets in those examples are the open sets of a topology, not its neighborhoods. $\endgroup$ Commented Aug 20, 2015 at 4:16

1 Answer 1


The empty set $\varnothing$ is not a neighborhood of any point $x\in X$, because as you correctly observed, there are no elements of $\varnothing$. However, the examples you are citing are not claiming that $\varnothing$ is a neighborhood, so there is no contradiction.

The examples are sets $X$ together with a topology $\tau$, so to say that a subset $U\subseteq X$ satisfies $U\in\tau$ means that $U$ is an open set, not (necessarily) a neighborhood of any particular point $x$.

From that Wikipedia page:

Given such a structure, we can define a subset $U$ of $X$ to be open if $U$ is a neighbourhood of all points in $U$.

The subset $U=\varnothing$ vacuously satisfies this property, because there are no points in $\varnothing$.


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