# Proof for convergent sequences, limits, and closed sets?

• Eclair_de_XII
In summary, this problem is asking you to show that a set, E, is closed if there exists a sequence of points in E that converges to a given point. You assume that this point is in E, and then show that E is closed.
Eclair_de_XII

## Homework Statement

"Let ##E \subset ℝ##. Prove that ##E## is closed if for each ##x_0##, there exists a sequence of ##x_n \in E## that converges to ##x_0##, it is true that ##x_0\in E##. In other words, prove that ##E## is closed if it contains every limit of sequences for each of its elements."

2. Homework Equations
Closed: A set is closed if it contains all of its accumulation points.

## The Attempt at a Solution

Choose an arbitrary ##x \in E##. Then there exists a sequence ##\{x_n\}_{n=1}^\infty## that converges to ##x##, where ##x_n\in E,\forall n\in ℕ##. Let ##\epsilon>0##. Then there exists an ##N\in ℕ## such that if ##n\geq N##, then ##|x_n - x|<\epsilon##. Equivalently, for ##n\geq N##, ##x_n\in (x-\epsilon,x+\epsilon)##. This neighborhood of ##x## contains all but finitely many ##x_n\in E##. Moreover, any neighborhood of ##x## contains infinitely many ##x_n \in E##. Therefore, ##x## is an accumulation point of ##E##. Since ##x\in E##, then it is implied that any ##x \in E## is an accumulation point of ##E##. The set ##E## is closed, as a result, since it contains all of its accumulation points.

I felt like this proof was too easy. Is there something I'm missing?

Eclair_de_XII said:

## Homework Statement

"Let ##E \subset ℝ##. Prove that ##E## is closed if for each ##x_0##, there exists a sequence of ##x_n \in E## that converges to ##x_0##, it is true that ##x_0\in E##. In other words, prove that ##E## is closed if it contains every limit of sequences for each of its elements."

2. Homework Equations
Closed: A set is closed if it contains all of its accumulation points.

## The Attempt at a Solution

Choose an arbitrary ##x \in E##. Then there exists a sequence ##\{x_n\}_{n=1}^\infty## that converges to ##x##, where ##x_n\in E,\forall n\in ℕ##. Let ##\epsilon>0##. Then there exists an ##N\in ℕ## such that if ##n\geq N##, then ##|x_n - x|<\epsilon##. Equivalently, for ##n\geq N##, ##x_n\in (x-\epsilon,x+\epsilon)##. This neighborhood of ##x## contains all but finitely many ##x_n\in E##. Moreover, any neighborhood of ##x## contains infinitely many ##x_n \in E##. Therefore, ##x## is an accumulation point of ##E##. Since ##x\in E##, then it is implied that any ##x \in E## is an accumulation point of ##E##. The set ##E## is closed, as a result, since it contains all of its accumulation points.

I felt like this proof was too easy. Is there something I'm missing?

In what set is ##x_0## in the original problem statement?

Is this an exact quote? I'm bothered by the part in red, which seems like an inadvertent add-on.
Eclair_de_XII said:
"Let ##E \subset ℝ##. Prove that ##E## is closed if for each ##x_0##, there exists a sequence of ##x_n \in E## that converges to ##x_0##, it is true that ##x_0\in E##.

Your proof is wrong. I did not look at the details, but you started with a point in E and then showed that it is an accumulation point.

But, who says that such a point is in E? For example, 0 is an accumulation point of ##\{1/n : n \in N_0\}##

What definition of accumulation point do you use?

The second sentence in your problem statement isn't the same as in the image.
Eclair de XII said:
"Let ##E \subset ℝ##. Prove that ##E## is closed if for each ##x_0##, there exists a sequence of ##x_n \in E## that converges to ##x_0##, it is true that ##x_0\in E##."

From the image, it is
"Let ##E \subset ℝ##. Prove that ##E## is closed if for each ##x_0## such that there is a sequence ##\{x_n\}_{n=1}^\infty## of points of E converging to ##x_0##, it is true that ##x_0 \in E##.

Math_QED said:
What definition of accumulation point do you use?

Accumulation point: A point ##x## is an accumulation point of ##E \subseteq ℝ## if every neighborhood of ##x## contains infinitely many elements of ##E##.

Mark44 said:
"Let ##E\subset ℝ##. Prove that ##E## is closed if for each ##x_0## such that there is a sequence ##\{x_n\}_{n=1}^\infty## of points of ##E## converging to ##x_0##, it is true that ##x_0\in E##.

Then I'm a bit confused on how to prove this, then. I think that I probably shouldn't assume ##x_0\in E##. But the second sentence is kind of confusing to me.

Last edited:
Eclair_de_XII said:
Accumulation point: A point ##x## is an accumulation point of ##E \subseteq ℝ## if every neighborhood of ##x## contains infinitely many elements of ##E##.
Mark44 said:
"Let ##E\subset ℝ##. Prove that ##E## is closed if for each ##x_0## such that there is a sequence ##\{x_n\}_{n=1}^\infty## of points of ##E## converging to ##x_0##, it is true that ##x_0\in E##.
Then I'm a bit confused on how to prove this, then. I think that I probably shouldn't assume ##x_0\in E##. But the second sentence is kind of confusing to me.
It is worded kind of odd, I agree. I believe you can assume that ##x_0## is in E, and that for each ##x_0## there is a sequence that converges to ##x_0##. You need to then show that E is closed. I don't think this is very difficult, as each ##x_0## is a limit point for some sequence.

Even so, I don't think it's a good idea to let ##x_0\in E## at first. Because if I define a sequence ##\{x_n\}_{n=1}^\infty## that converges to ##x_0##, it may be a constant sequence ##\{x_n:x_n=x_0,n\in ℕ\}## and I would be unable to prove that ##x_0## is an accumulation point.

Mark44 said:
It is worded kind of odd, I agree. I believe you can assume that ##x_0## is in E, and that for each ##x_0## there is a sequence that converges to ##x_0##. You need to then show that E is closed. I don't think this is very difficult, as each ##x_0## is a limit point for some sequence.

I'm pretty sure that ##x_0## mustn't be necessarily in ##E## (see my comment above), but I agree that the question is stated terribly.

Let us begin, now we know what has to be proven and what an accumulation point is:

Denote ##\overline{E} := \{x \in \mathbb{R} \mid \forall \epsilon > 0: (x- \epsilon, x + \epsilon) \cap E \mathrm{\ contains \ infinitely \ many \ points} \}## with the set of all accumulation points of ##E##.

We have to prove that ##\overline{E} \subseteq E##. The other inclusion is always true (can you see why?)

For this, I will provide you with a strategy to tackle this problem:

Step 1: Take ##x \in \overline{E}##
Step 2: Show that there is a sequence in ##E## that converges to ##x##.
Step 3: Conclude that ##x \in E##, using the assumption in the problem.
Step 4: Conclude that ##E = \overline{E}##

Gotcha. Thanks, everyone. I'll try my best on the problem, then.

## 1. What is a convergent sequence?

A convergent sequence is a sequence of real numbers that approaches a finite limit as the number of terms increases. In other words, as the sequence continues, the numbers get closer and closer to a specific value.

## 2. How do you prove a sequence is convergent?

To prove a sequence is convergent, you must show that the terms of the sequence eventually get arbitrarily close to the limit value. This can be done by using the definition of convergence, which states that for any positive number, there is a point in the sequence where all subsequent terms will be within that distance from the limit.

## 3. What is a limit of a sequence?

The limit of a sequence is the value that the terms of the sequence approach as the number of terms increases. It is not necessarily a term in the sequence, but rather the value that the terms converge to.

## 4. How do you calculate the limit of a sequence?

To calculate the limit of a sequence, you can use various methods such as the squeeze theorem, the ratio test, or the root test. These methods involve manipulating the terms of the sequence to determine if they approach a specific value or if the sequence diverges.

## 5. What is a closed set?

A closed set is a set that contains all its limit points. In other words, every sequence of points in the set that converges must have its limit point also in the set. This is often represented by the closure of the set, denoted by a bar over the set symbol.

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