8.1 Neighbourhood, adherent point, isolated point, accumulation point[29V]
[0GW] 1 Let \((X, τ )\) be a topological space and let \(x_ 0 ∈ X\).
We denote as neighbourhood of \(x_ 0\) any superset of an open set containing \(x_ 0\) .
We call fundamental system of neighbourhoods of \(x_ 0\) a family \(\{ U_ i \} _{i∈I}\) of neighborhoods \(x_ 0\) with the property that each neighborhood of \(x_ 0\) contains at least one of the \(U_ i\) .
We will say that \(U\) is an open neighborhood of \(x_ 0\) simply to say that \(U\) is an open set that contains \(x_ 0\).
[0GX] Let \(E,F⊆ X\) be sets:
a point \(x_ 0 ∈ X\) is an adherent point of \(E\) if every neighborhood \(U\) of \(x_ 0\) has non-empty intersection with \(E\);
a point \(x_ 0 ∈ E\) is isolated in \(E\) if there exists a neighborhood \(U\) of \(x_ 0\) such that \(E ∩ U = \{ x_ 0 \} \);
(Note that, in some cases, sets can have at most a countable number of isolated points: see 2 and 1, and also 2).
We also define this concept (already seen in 176 for the case \(X={\mathbb {R}}\)).
- E250
[0GZ]Check that in the definitions 249 and 250 you can equivalently use, instead of the neighborhoods \(U\) of \(x_ 0\), the open neighborhoods \(U\) of \(x_ 0\).
- E250
[0H0] Check that in the definitions 249 and 250 you can equivalently use neighborhoods \(U\) of \(x_ 0\) chosen from a fixed fundamental system of neighborhoods.
- E250
[0H1] Check that the set of points adhering to \(A\) coincides with the closure of \(A\). Hidden solution: [UNACCESSIBLE UUID ’0H2’]
- E250
[0H3] Prerequisites:3.Check that \(\overline A=A∪ D(A)\). Hidden solution: [UNACCESSIBLE UUID ’0H4’]
- E250
[0H5] A point \(x\in X\) is an accumulation point for \(X\) 4 if and only if the singleton \(\{ x\} \) is not open. Hidden solution: [UNACCESSIBLE UUID ’0H6’]
- E250
[0H7] Topics:boundary. Let \(A⊂ X\). Let’s remember the definition of boundary \(∂ A=\overline A⧵ {{A}^\circ }\). Note that \(∂ A\) is closed: indeed setting \(B=A^ c\) to be the complement, it is easily verified that \(∂ A=\overline A∩ \overline B\). In particular we showed that \(∂ A=∂ B\).
Show that the three sets \(∂ A,{{A}^\circ },{{B}^\circ }\) are disjoint, and that their union is \(X\); in particular, show that the three sets are characterized by these three properties:
Each neighborhood of \(x\) intersects both \(A\) and \(B\);
there exists a neighborhood \(x\) contained in \(A\);
there exists a neighborhood \(x\) contained in \(B\).
(See also 286 for the case of metric spaces). Hidden solution: [UNACCESSIBLE UUID ’0H8’]
- E250
[0H9] Topics:boundary.Difficulty:*.
Given \(X\) topological space and \(A⊆ X\); if \(A\) is open (or closed) the boundary \(∂ A\) has empty interior; we have \(∂ A⊇ ∂∂ A\) with equality if \(∂ A\) has empty interior; in addition \(∂∂ A= ∂∂∂ A\). Hidden solution: [UNACCESSIBLE UUID ’0HB’][UNACCESSIBLE UUID ’0HC’]
- E250
[0HD] Prerequisites:4.If \((X,𝜏)\) is a topological space and \(A⊂ X\) has no isolated points, then also \(\overline A\) does not have isolated points. Hidden solution: [UNACCESSIBLE UUID ’0HF’]
- E250
[0HG]Note:written exam, 12/1/2013.Let A be an open subset of \(X\). Prove that, for any subset B of \(X\) , the following inclusion holds: \(A ∩ \overline B ⊆ \overline{A ∩ B}\). Show, by an example, that the conclusion does not hold if you remove the assumption that A is open. Hidden solution: [UNACCESSIBLE UUID ’0HH’]
- E250
[0HJ]Given \(E⊆ X\), we distinguish the points \(x∈ X\) in three distinct sets that are a partition of \(X\).
For every neighborhood \(U\) of \(x\), \(U⧵\{ x\} \) intersects \(E\). These are the accumulation points of \(E\).
\(x∈ E\) and there is a neighborhood \(U\) of \(x\) such that \(U∩ E =\{ x\} \). These are the isolated points in \(E\).
Now describe the third set of points
Hidden solution: [UNACCESSIBLE UUID ’0HK’]