Since "neighborhoods" can be defined differently (as noted in the comments to this question), here are the relevant definitions I'm working with:

Topology (defined via open sets)

(from wiki; slightly modified for clarity)

A topology on a set $X$ may be defined as a collection $\tau$ of subsets of $X\text{,}$ satisfying the following axioms:

  1. The empty set and the carrier set belong to the topology. That is, $\varnothing \in \tau$ $\text{and } X \in \tau \text{.}$
  2. Any arbitrary (finite or infinite) union of members of $\tau$ belongs $\text{to }\tau \text{.}$
  3. The intersection of any finite number of members of $\tau$ belongs $\text{to }\tau \text{.}$

The carrier set, along with its topology, is called a topological space and is denoted $\text{by }(X, \tau)\text{.}$

Open set

Any element in the topology is called an open set. That is, any set $U$ $\text{where }U \in \tau \text{.}$

Neighbourhood (of a point)

(from wiki; slightly modified for clarity)

If $(X, \tau)$ is a topological space and $p$ is a point in $X$, then a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ $\text{containing }p$,

$$p \in U \subseteq V \subseteq X\text{.}$$

The Question

What are some examples of open sets that are not neighborhoods? The only one that comes to my mind is this one:

  • The empty set $\varnothing$ since it's open and contains no points of $X$, regardless of the topology.

I tried thinking about the discrete topology on $X\text{,}$ and briefly believed that singleton sets were an example (until I started writing out my train of logic on this post). My first line of thinking was something like this:

Every point $p \in X$ has a corresponding singleton set $\{p\} \in \tau$ (so it's open). However, $\{p\}$ is the smallest open set containing the point $p\text{,}$ so there doesn't exist any $U$ so that $$p \in U \subseteq \{p\} \subseteq X\text{.}$$

But then I remembered that any set is a subset of itself, so $\{p\} \subseteq \{p\}$ made me realize that the sets $U$ and $V$ in the definition of neighborhood could be the same set.

I guess what I'm really wondering is whether there are non-empty open sets (in some topology) that are not neighborhoods.

  • 1
    $\begingroup$ I think the OP's confusion stems from trying to make sense of a concept of "neighbourhood" as something distinct from "open set", when in reality the only meaningful use of the word neighbourhood is when specifying "neighbourhood of a particular point p." $\endgroup$ Commented Jul 5, 2023 at 18:16
  • $\begingroup$ Slight nitpick: one can also have a "neighbourhood of a set" en.wikipedia.org/wiki/… But you made me realize that I shouldn't have left it implicit for most of the post (figured the section title "Neighbourhood (of a point)" was enough 🤷). A part of my confusion likely stems from some mathematicians using "U is a neighborhood of a point p" to mean "U is an open set containing p." (p. 96 of Munkres Topology 2nd edition) But I think you're also correct that another part of my confusion was from not focusing on a point p. $\endgroup$ Commented Jul 6, 2023 at 5:00

1 Answer 1


An easy consequence of the definition is that an open set is a neighborhood of each of its members. Namely, if $U$ is open and $p \in U$, take $V = U$ in the definition of neighborhood.

  • $\begingroup$ Never mind, I see what you're saying! $\endgroup$
    – Brian Tung
    Commented Jul 5, 2023 at 2:42
  • 2
    $\begingroup$ Are you not qualifying "an open set" with "a non-empty open set" because "an open set is a neighborhood of each of its members" is vacuously true when the open set is empty? $\endgroup$ Commented Jul 5, 2023 at 2:50
  • 1
    $\begingroup$ @KevinFlowersJr, actually when Robert says “open set”, he means “non-empty open set”, indeed it is obvious that the empty set is open but cannot be a neighbourhood given that it does not contain any point. $\endgroup$
    – Angelo
    Commented Jul 5, 2023 at 8:17
  • 4
    $\begingroup$ @Angelo: …and yet, the empty set is indeed obviously (and vacuously) "a neighborhood of each of its members" (of which there are none)! $\endgroup$ Commented Jul 5, 2023 at 11:18
  • 7
    $\begingroup$ @Angelo@ Robert is not claiming that the empty set is a neighbourhood. It is, however, a neighbourhood of each of its points (because it has no points). $\endgroup$
    – TonyK
    Commented Jul 5, 2023 at 12:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .