# Theorem:

Let M be a nonempty subset of a metric space (X, d). $$\overline{M}$$ is M's closure then:

$$(1)x \in \overline{M} \iff \exists \{x_i\}_{i=1}^{n} \in M, x_n \to x$$

### Question 1: Can we make statement (2) from statement (1)

$$(2)x \in \overline{M} \iff \exists \{x_i\}_{i=1}^{n} \in \overline{M}, x_n \to x$$

# Proof:

## proof 1

Let $$x \in \overline{M}$$

### Question 2: This means x is in $$M$$ or in $$\overline{M}$$, right?

Let $$x \in M$$, then we have a sequence $$(x,...,x)$$

### Question 3: Does this answer Question 1 ?

Let $$x \notin M$$, Then for every $$n=1,2,...$$ we can have a ball $$B(x,\frac{1}{n})$$, which contains $$x_{n} \in M$$

and $$x_{n} \to x$$

### Question 4 Why ?

Question 6 How does proving a statement for a specific ball $$B(x,\frac{1}{n})$$ proves the statement in general ?

## Proof for <=

### Question 5 What is the idea?

• Half of your 1st displayed line is MISSING. Please edit. Otherwise nobody can tell what the Q is. Aug 7, 2019 at 22:10
• I corrected your paraphrasing so that it is correct: $x$ is in the closure iff there exists a sequence …. such that ... Aug 7, 2019 at 23:32
• @DanielWainfleet I am sorry, edit has been made. Aug 8, 2019 at 7:15
• @DisintegratingByParts I have made a proper edit which asks what I want to ask :) Aug 8, 2019 at 7:15

The correct version of (1) is $$x \in cl(M) \Leftrightarrow \hspace{.2cm} \exists \{x_n\}_{n=1}^\infty with \hspace{.2cm} x_n \in M \hspace{.2cm} \forall n \in \mathbb{N} \hspace{.2cm} \wedge x_n \overset{n \rightarrow \infty} {\rightarrow} x.$$

As you point out if $$x \in M$$ the sequence $$x_n = x \hspace{.2cm} \forall n \in \mathbb{N}$$ makes x an element of cl(M).

The claim you give in Q3 is also a correct proof you only interpret it inaccurately.

Since you in essence say $$\forall n \in \mathbb{N} \hspace{.2cm} \exists x_n \in B(x, \frac{1}{n}),$$ you can now define the sequence $$\{x_{n_k}\}_{k=1}^\infty,$$ where you select the elements of the sequence in the following way:

k = 1 pick any point in $$M \backslash \{x\}$$ to be $$x_{n_1}$$ and compute $$d(x_1,x) = d_1$$. $$\forall k > 1$$ pick $$x_{n_k} \in B(x,r_k)$$, where $$r_k = max(\{\frac{1}{n}| \frac{1}{n} < d_{k-1}\})$$. This guarantees that $$l > k \Rightarrow d_l < d_k$$. By the Monotone sequences Theorem this means that $$d_k \overset {k\rightarrow \infty}{\rightarrow} 0,$$ since $$d_k \leq 0$$ by definition. I.e $$x_{n_k} \overset {k\rightarrow \infty}{\rightarrow} x.$$

With respect to Q5 $$\forall r \in \mathbb{R} \hspace{.2cm} \exists k \in \mathbb{N} \hspace{.2cm} : r_k < r \hspace{.2cm} \Rightarrow x_{n_k} \in B(x,r),$$ which proves the claim.

• +1... Statement (2) in the Q also needs correcting. Aug 10, 2019 at 23:19