Proof for convergent sequences, limits, and closed sets?

[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?

 

[member 587159]

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?

 

[Mark44]

Is this an exact quote? I'm bothered by the part in red, which seems like an inadvertent add-on.

"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 $x_0\in E$.

 

[Eclair_de_XII]

Sorry for the late reply... Anyway, here's the original text: https://i.imgur.com/7OVX0Re.png

 

[member 587159]

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?

 

[Mark44]

The second sentence in your problem statement isn't the same as in the image.

"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 $x_0 \in E$.

 

[Eclair_de_XII]

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 $E$.

"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.

 

[Mark44]

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

"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.

 

[Eclair_de_XII]

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.

 

[member 587159]

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.

 

[member 587159]

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 $E = \overline{E}$

 

[Eclair_de_XII]

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