X in Boundary => x Isolated or Strict Limit (topological space)

In summary, the proposition states that if an element x is in the boundary of a set A in a topological space, then x is either an isolated point or a strict limit point of both A and the closure of A. The definitions used to understand this proposition include the closure of a set, isolated points, and strict limit points. The reasoning behind this is that if x is not isolated, then it is a strict limit point of A and the closure of A. However, this logic may not work if there are an uncountable number of open sets. To show that x is a strict limit point of A, a sequence can be constructed using open sets containing x and elements in A.
  • #1
NickAlger
15
0
Although the intuition makes sense, I am having trouble determining why the following proposition is correct. The document leaves this as an exercise for the reader... great.

Proposition: Suppose A is a set in a topological space, and dA is the boundary of A. If x is in dA, then x is either an isolated point, or a strict limit point of both A and A closure.

---

The definitions I am working from are:
1) dA is (A closure)∩((A complement) closure)

2) x in A is isolated if there is an open set E containing x such that E\{x} does not intersect A. x is sooo lonely, surrounded by strangers... ;)

3) x is a strict limit point of A if there is a sequence {x_n} contained in A\{x} such that x_n -> x.

4) x_n -> x if every open set containing x also contains the tail of the sequence {x_n}.

---

So far, my reasoning is this. Supposing x is not isolated, let's show that it is a strict limit point of A by constructing a sequence {x_n} in A that converges to x. After this, showing it for A closure will just requiring twiddling a few words.

If the number of open sets containing x were countable, {E1, E2, E3, ...}, then I would put x1 in E1\{x}, x2 in (E1 intersect E2)\{x}, x3 in (E1 intersect E2 intersect E3)\{x}, and so forth.

We know these intersections can't be empty - if Ei and Ej are open and contain x, then (Ei intersect Ej) is also an open set containing x. Since x is not isolated, (Ei intersect Ej) must contain some y in A (distinct from x). Thus for a countable number of open sets, we can construct the proper convergent sequence.

However, this logic will not work if there are an uncountable number of open sets. I might be able to extend my argument if only the basis is countable, but we don't know that. For example, in the real numbers centered at 0, you might start making up your sequence based on a particular countable collection of open sets like {Ball(.51), Ball(.501), Ball(.5001), etc} and never get to any of the "smaller" open sets.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
##x## not isolated means: every open set ##U## containing ##x## has a second element ##x\neq a \in A##.
We further know that ##x \in \partial A = A^{-}\cap (X\backslash A)^{-}##.
You didn't give us your definition of closure, so I assume we can say, that the closure of a set is the set itself plus all its limit points, strict or not. Since ##x\in A## we have ##x \notin X\backslash A##. Hence it is a strict limit point of ##X\backslash A##, i.e. there is a sequence ##y_n \longrightarrow x## in ##X\backslash A##. This means any open set containing ##x##, contains the tail of this sequence, too.

We want to show that ##x## is a strict limit point of ##A##, i.e. that there is a sequence ##x_n \longrightarrow x## in ##A\backslash \{\,x\,\}##, i.e. any open set which contains ##x## must contain the tail of such a sequence.

Let ##x\in U## be any open set. Now we build a sequence of open sets ##U_n## defined by ##x\in U_n ## and ##y_k \in U\cap U_n## for all ##k \geq n##. Every ##U\cap U_n## contains all elements ##x\neq a_k \in A## with ##k\ge n##, hence ##x## is a strict limit point of ##A##.
 

1. What is a topological space?

A topological space is a mathematical concept used in the field of topology to study the properties of spaces that are preserved under continuous deformations. It consists of a set of points and a collection of open sets, which are subsets of the space that satisfy certain axioms.

2. What does it mean for a point to be isolated in a topological space?

A point in a topological space is considered isolated if it is not a limit point of any subset of the space. This means that the point is not surrounded by any other points in the space, and therefore has a neighborhood that does not contain any other points.

3. How is a strict limit different from an isolated point in a topological space?

A strict limit of a point in a topological space is a point that is surrounded by infinitely many other points in the space. This means that every neighborhood of the point contains points other than the point itself. In contrast, an isolated point does not have any other points in its neighborhood.

4. Can a topological space have both isolated points and strict limits?

Yes, a topological space can have both isolated points and strict limits. In fact, most topological spaces have both types of points. For example, in the real line, every point is both an isolated point and a strict limit point.

5. How does the concept of isolated points and strict limits relate to continuity?

In topology, continuity is a fundamental concept that describes the behavior of functions between topological spaces. Intuitively, a function is continuous if small changes in the input result in small changes in the output. The presence of isolated points and strict limits in a topological space can affect the continuity of a function defined on that space.

Similar threads

  • Calculus
Replies
3
Views
996
Replies
7
Views
2K
  • Topology and Analysis
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Replies
16
Views
9K
Replies
3
Views
781
Replies
3
Views
2K
  • Topology and Analysis
Replies
7
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Back
Top