The Geometric Viewpoint

Constant Sequences With Infinitely Many Limits

Posted on November 18, 2014 1:23 pm by sliukase formatted with Image Comment

Back in calculus, we were taught that any convergent sequence ( ${x_n}$ ) in ${\mathbb{R}}$ has exactly one limit. This statement is true, and you might have seen the proof before. But is this necessarily a case in non- ${\mathbb{R}}$ spaces?

It turns out that this is still true in some non- ${\mathbb{R}}$ spaces, but, as we will see later, not all.

But what are spaces where things behave strangely? First, let’s consider the following definition:

Definition: a space H is Hausdorff if for any two distinct points you pick, there are disjoint open sets containing them.

An example of a Hausdorff space is ${\mathbb{R}}$ (the real line). By a standard fact, we know that for any two distinct real numbers x and y, we can find disjoint open intervals containing them (another way to think of this: suppose that the distance between x and y is d. Start by letting ${{I_x}}$ be an open interval centered at x with radius d/2 and let ${{I_y}}$ be an open interval centered at y with radius d/2 as well. It is easy to observe that these two intervals are disjoint.).

It turns out that in every Hausdorff space, a convergent sequence has exactly one limit (as we will prove later). On the other hand, in a non-Hausdorff space, a constant sequence (which obviously converges) can have infinitely many limits!

It might be hard to image what non-Hausdorff spaces look like, so we will begin by introducing the concept of topological spaces.

Given a set X, (X, T) is a topological space exactly when these four conditions hold:

1. X ${\in}$ T

2. ${\emptyset}$ ${\in}$ T

3. If a finite number of sets are elements of T, then their intersection ${\in}$ T

4. If we have sets that are elements of T, then their union ${\in}$ T

and T is called a topology on X (don’t confuse it with topography). Elements of T are exactly open sets in (X, T).

Observation: A set can be both closed and open. By the definition, C is a closed set exactly when the complement of C is open. For any given set X and topology T on X, since X and ${\emptyset}$ are in T, they are open sets in the space (X, T). Thus, their complements, ${\emptyset^c}$ = X and ${X^c}$ = ${\emptyset}$ are both closed.

The picture below represents a (X, T) topological space, where X = {1,2} and T = {ø, {1},{2},{1,2}}. It is easy to check that this is a Hausdorff space.

Before we move on to examples of sequences in non-Hausdorff topological spaces with multiple limits, let’s recall the definition of a limit of a sequence.

Definition: x is a limit of sequence ( ${x_n}$ ) if and only if for any open set U containing x, eventually there is a term of ( ${x_n}$ ) in U such that all the succeeding terms of ( ${x_n}$ ) are in U as well.

2 Examples are given here:

Example 1:

Let F be {1, 2, 3} and T = { ${\emptyset}$ , {1, 2}, F}. T satisfies all the 4 conditions, so it is a topology on set F. Obviously, (F, T) is not Hausdorff. Consider 1 and 2. Open sets containing 1 are exactly those that contain 2. Thus, regardless of your choice of the pair of open sets containing 1 and 2, the intersection is never empty.

Let ( ${x_n}$ ) be a constant sequence such that ${x_n}$ = 1 for every n. For any open set containing 1 (there are 2 of them: ({1,2} and F), every term of ( ${x_n}$ ) is in the open set, thus, ( ${x_n}$ ) converges to 1. On the other hand, since open sets containing 1 are exactly ones that contain 2, ( ${x_n}$ ) also converges to 2. It is easy to see that the open set containing 3 also contains all terms of ( ${x_n}$ ). Thus, the set of limits of ( ${x_n}$ ) is {1, 2, 3}.

Example 2:

Let’s consider a more interesting example. Let X = ${\mathbb{N}}$ (the set of all natural numbers), and let T = { ${\emptyset}$ , X}. Obviously, T is a topology on X ( ${\textbf{note:}}$ for any set X, T = { ${\emptyset}$ , X} is called an indiscrete topology on X). Clearly, this is not Hausdorff.

Let ( ${x_n}$ ) be a constant sequence such that ${x_n}$ = 1 for every n. Let j be an arbitrary natural number. Since every open set in the space (X, T) containing j contains all terms of ( ${x_n}$ ), j is a limit of ( ${x_n}$ ). Thus, every natural number is a limit point of ( ${x_n}$ ) in (X, T). This points out that the set of limit points does not even need to be finite!

Intuitively, the result of this example may seem contradictory, since the given sequence is constant but is also “arbitrarily” close to infinitely many points. This shows that things could behave bizarrely in non-Hausdorff spaces.

Convergent sequences and their limits in Hausdorff spaces

After seeing these two examples, one might wonder if a convergent sequence in a Hausdorff topological space always has exactly one limit. It turns out that this is true.

Let (X, T) be a Hausdorff topological space and ( ${x_n}$ ) be a convergent sequence in the space, we’ll show that it has exactly one limit.

We’ll prove our claim by contradiction. In other words, we’ll assume that our claim is false and show that such assumption can’t be true.

For contradiction, suppose that ( ${x_n}$ ) has two distinct limits x and y in X. Since the space is Hausdorff, there are open sets ${U_x}$ and ${U_y}$ containing x and y respectively, and the intersection ${U_x}$ ${\cap}$ ${U_y}$ = ${\emptyset}$ Since ( ${x_n}$ ) converges to x, there is ${x_{n_k}}$ ${\in}$ ${U_x}$ such that all the suceeding ${x_n}$ are in ${U_x}$ .

Similarly, since ( ${x_n}$ ) converges to y, there is ${x_{n_m}}$ ${\in}$ ${U_y}$ such that all the suceeding ${x_n}$ are in ${U_y}$ . There are 2 cases:

case1: ${n_{k}}$ ${\geq}$ ${n_{m}}$ ${\longrightarrow}$ ${x_{n_k}}$ is in both ${U_x}$ and ${U_y}$

case2: ${n_{k}}$ ${\textless}$ ${n_{m}}$ ${\longrightarrow}$ ${x_{n_m}}$ is in both ${U_x}$ and ${U_y}$

Thus, ${U_x}$ ${\cap}$ ${U_y}$ ${\neq}$ ${\emptyset}$ , which is a contradiction.

This points out that our assumption is false, and ( ${x_n}$ ) must have exactly one limit in X

Since the space and the sequence are arbitrary, any convergent sequence in a Hausdorff space has exactly one limit.

Convergent sequences and their limits in “weakened” Hausdorff spaces

In mathematics, changing hypothesis can drastically change the result, and the idea that we are exploring here is not an exception. Although a convergent sequence in any Hausdorff space has the unique limit, a convergent sequence in a “weaker” Hausdorff space can have many limits!

Definition: a space H is T1 if for any two distinct points, each has a neighborhood which does not contain the other point.

We can observe that this is weaker than the definition of a Hausdorff space, since the neighborhoods are not required to be disjoint. If for each pair of points, there are neighborhoods that happen to be disjoint, then this T1 space is also a Hausdorff space. However, that does not have to be a case. Obviously, every Hausdorff space is also T1, but not the other way around.

But how does using a weaker condition affect the behavior of convergent sequences? As we have seen, a convergent sequence in any Hausdorff space has exactly one limit, and a convergent sequence in a non-Hausdorff space may have many limits, but what can we say about a convergent sequence in a T1 space?

It turns out that it is possible to find a T1 space where a convergent sequence has infinitely many limits! Let’s consider the following example: space (X, T) when X = ${\mathbb{N}}$ and T = { ${O}$ $\subset$ ${\mathbb{N}}$ | ${O^c}$ is finite} $\cup$ { $\emptyset$ }

Proving that this is a topological space is not hard, and we will skip it to discuss more interesting results.

First, we will show that (X, T) is a T1 space. Let a,b be distinct points of X. Since X\{a} and X\{b} have finite complements, they are elements of T. Given that X\{a} contains b but not a and X\{b} contains a but not b, (X, T) is a T1 space.

Consider a sequence ( ${x_n}$ ) in (X, T) where ${x_n}$ = n for each n ${\in}$ ${\mathbb{N}}$ . We will show that every natural number is a limit point of ( ${x_n}$ ). Let j be an arbitrary natural number and let ${U_j}$ be an open set containing j. Since ( ${U_j}$ ) is an element of T, the complement of ( ${U_j}$ ) is finite. In other words, C( ${U_j}$ ) is either ${\emptyset}$ or { ${n_1}$ , ${n_2}$ , …, ${n_k}$ } (where these ${n_j}$ are natural numbers). Without loss of generality, we can assume that, in the second case, ${n_k}$ is the biggest element of such finite set.

If C( ${U_j}$ ) is ${\emptyset}$ , then ${U_j}$ contains all the natural numbers. Thus, it contains every term of the sequence ( ${x_n}$ ).

On the other hand, if If C( ${U_j}$ ) is { ${n_1}$ , ${n_2}$ , …, ${n_k}$ }, then ${U_j}$ contains all n > ${n_k}$ . Equivalently, for all n > ${n_k}$ , ${x_n}$ is contained in ${U_j}$ .

${U_j}$ is arbitrary, so ( ${x_n}$ ) converges to j.

But j is also arbitrary, so the sequence ( ${x_n}$ ) converges to all natural numbers.

Question:

We’ve seen that every convergent sequence in any Hausdorff space has exactly one limit and that there are non-Hausdorff spaces where a convergent sequence can have more than one limit. Is it possible to have a convergent sequence in a non-Hausdorff space with exactly one limit?

Saran was a student in Scott Taylor’s Fall 2014 Topology course at Colby College.

References

http://math.stackexchange.com/questions/901154/unique-limits-in-t1-spaces

http://en.wikipedia.org/wiki/Interior_(topology)#mediaviewer/File:Interior_illustration.svg

https://dragonflytraining.files.wordpress.com/2013/10/man-with-question-01.png

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