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The definition of basis and neighborhood basis are:

Let $(X,\tau)$ be a topological space, a base of $\tau$ is a subset $\mathfrak{B}$ of $\tau$ such that each open set $A \in \tau$ is union of elements of $\mathfrak{B}$

If $p \in X$, a subset $\mathfrak{B}_p\subseteq U_p=\{U \in \tau | p \in U\}$ of neighborboods of $ p$ is called a neighborhood basis of $ p$ if for each $U \in U_p$ there exist an $V\in \mathfrak{B}_p$ such that $V\subseteq U$

I have the following example in my lecture notes:

If X is a set such that $|X|>\aleph_0$ and $\tau$ is the discrete topology on $X$, now $(X,\tau)$ does not have a countable basis. In fact, let $\mathfrak{B}$ be a basis of $\tau$: Because $\{p\} \in \tau$ for each $p \in X$, the set $\{p\}$ must be union of elements of $\mathfrak{B}$. Therefore $\{p\}\in \mathfrak{B}$ and $|\mathfrak{B}|\geq|X|>\aleph_0$.

Furthermore, each neighborhood of $p \in X$ contains the open set $\{p\}$, so it is a finite neighborhood basis of $p$

I don't understand quite well the concept of neighborhood basis so to understand the last paragraph, I came up with a concrete example: Let $X=\{1,2,3\}$ and $(X,\tau)$ a topological space with discrete topology.

Then $\tau$ equals the power set $\tau=\{ \{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},X,\emptyset \}$

If I choose $p=1$ the set of neighborhoods of $p$ are: $U_p= \{ \{1\},\{1,2\},\{1,3\},X \}$

so that $\mathfrak{B}_p \subseteq U_p$ , $\mathfrak{B}_p=\{\{1\}\}$ is a neighborhood basis because it is contained in each element of $U_p$ in agreement with the definition of neighborhood basis

and

$\mathfrak{B} \subseteq \tau$ ,$\mathfrak{B}=\{ \{1\},\{2\},\{3\} \}$ is a basis

  1. Is this example correct? and does it correctly explain why $\{p\}$ is a finite neighborhood basis of $p$ in the initial example?

  2. Assuming this example is correct, just like when having a basis, each element of the topology can be expressed as union of elements of the basis, I was expecting that when having a neighborhood basis, each element of the set of neighborhoods of a point $p$ could be expressed as union of elements of the neighborhood basis, but it looks like it is not the case, since with the basis $\mathfrak{B}_p=\{\{1\}\}$ , I can't express the elements$ \{1,2\},\{1,3\}$and $X $ of $U_p$ as union of elements of $\mathfrak{B}_p$. How do I make sense of this?

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2 Answers 2

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Yes, your example is correct.

The basis and the neighbourhood basis, despite the similar name have two different scopes.

The idea behind the basis of the topology is that you want something simpler from which it is possible to reconstruct the full topology.

The neighbourhood basis does not help you in finding all the possible neighbourhoods of a point. It's quite the opposite goal. It wants to select a subset of neighbourhoods which allows to study the property of what's happening around this point only. So it's a set of neighbourhoods which can be finer and finer, and the finer they become the more they forget about the rest of the topology, which is not of interest when studying some property of a single point.

In your discrete example, when you look at the property of the element $1$ you may want to forget about all the rest. That's when the neighbourhood basis is just $\{1\}$. In the discrete topology, all the other points are as if disconnected from $1$, so you don't want them around when studying properties of the point $1$ only.

If the topology is not discrete then things can be more interesting. If you want to study the topological properties of some object around a point $x_0\in\mathbb{R}$ you don't need to look at all the open set which contain $x_0$. Often it is enough to look at small intervals around it. Indeed, a neighbourhood basis around $x_0$ (in the Borel topology) can be a sequence of open shrinking interval which is collapsing on $x_0$.

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    $\begingroup$ This answer gave useful intuition, too. I wish I could accept more than one answer $\endgroup$ Jul 3, 2020 at 18:27
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Your example is correct in the sense that $\mathfrak{B}$ is a base for the discrete topology on $X$: a topology is always a base for itself. However, $\mathfrak{B}_0=\big\{\{1\},\{2\},\{3\}\big\}$ is also a base for $\tau$, because every member of $\tau$ is the union of some subset of $\mathfrak{B}_0$:

$$\begin{align*} X&=\bigcup\mathfrak{B}_0=\{1\}\cup\{2\}\cup\{3\}\\ \{1,2\}&=\bigcup\big\{\{1\},\{2\}\big\}=\{1\}\cup\{2\}\\ \{1,3\}&=\bigcup\big\{\{1\},\{3\}\big\}=\{1\}\cup\{3\}\\ \{2,3\}&=\bigcup\big\{\{2\},\{3\}\big\}=\{2\}\cup\{3\}\\ \{1\}&=\bigcup\big\{\{1\}\big\}=\{1\}\\ \{2\}&=\bigcup\big\{\{2\}\big\}=\{2\}\\ \{3\}&=\bigcup\big\{\{3\}\big\}=\{3\}\\ \varnothing&=\bigcup\varnothing \end{align*}$$

In fact, every subset of $\tau$ that contains $\{1\},\{2\}$, and $\{3\}$ is a base for $\tau$.

In this space the family of nbhds of $1$ is, as you say $\mathfrak{N}(1)=\big\{\{1\},\{1,2\},\{1,3\},X\big\}$. This means that in order for a family $\mathfrak{B}_1$ of open sets to be a nbhd base at $1$, $\mathfrak{B}_1$ must satisfy two conditions:

  • every member of $\mathfrak{B}_1$ must be an open set containing $1$, and
  • every open set containing $1$ must contain some member of $\mathfrak{B}_1$.

The collection $\big\{\{1\}\big\}$ satisfies both of these conditions; so does any other subset of $\mathfrak{N}(1)$ that includes $\{1\}$. Note, though, that $\{1\}$ absolutely has to belong to any nbhd base at $1$, because it’s the only open nbhd of $1$ that is contained in $\{1\}$.

The answer to your final question is simply that you must realize that your expectation is incorrect: nothing in the definition of neighborhood base at $p$ requires or implies that every open nbhd of $p$ is a union of members of a nbhd base at $p$. What is true is that if $U$ is an open set, $p\in U$, and $\mathfrak{B}_p$ is a nbhd base at $p$, then there must be a $B_{p,U}\in\mathfrak{B}_p$ such that $p\in B_{p,U}\subseteq U$. And we can do this at each point of $U$: for each $p\in U$ there is a $B_{p,U}\in\mathfrak{B}_p$ such that $p\in B_{p,U}\subseteq U$, from which it follows that

$$U=\bigcup_{p\in U}B_{p,U}\;.$$

That is, $U$ need not be the union of members of any single nbhd base $\mathfrak{B}_p$, but it is the union of members of the nbhd bases of its elements. And this shows that if for each $p\in X$ we have a nbhd base $\mathfrak{B}_p$, then $\bigcup_{p\in X}\mathfrak{B}_p$ is a base for the topology on $X$.

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  • $\begingroup$ I think the term "nbhd basis" or "nbhd base at" makes it very easy to missunderstand it the way I did, thinking that it must work the same as a basis for the topology. Not sure why they chose such a misleading name. $\endgroup$ Jul 3, 2020 at 18:16
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    $\begingroup$ @J.C.VegaO: Because it really is descriptive, once you realize what it’s actually describing: a nbhd base at $p$ is not intended to say anything about what the topology looks like at any point except $p$, and it says exactly what is required for a set to be a nbhd of $p$ — that it must contain a member of the nbhd base. These are the sets that define the topology at the point $p$. And many topologies are most easily defined by specifying a nbhd base at each point. Part of the problem, I suspect, is that you learned about bases before you learned about nbhd bases; ... $\endgroup$ Jul 3, 2020 at 18:23
  • $\begingroup$ ... had the topics been introduced in the opposite order, as is sometimes done, I suspect that the question wouldn’t have arisen. $\endgroup$ Jul 3, 2020 at 18:24
  • $\begingroup$ Yes, indead I learned first about bases $\endgroup$ Jul 3, 2020 at 18:27
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    $\begingroup$ @l'étudiant: Yes. If $\langle X,\tau\rangle$ is a topological space, $\mathscr{B}$ is a base for $\tau$, and $x\in X$, then $\{B\in\mathscr{B}:x\in B\}$ is a nbhd base at $x$. And if for each $x\in X$ we have a nbhd base $\mathscr{B}_x$ at $x$, then $\bigcup_{x\in X}\mathscr{B}_x$ is a base for $\tau$. $\endgroup$ Mar 19 at 21:45

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