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 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?

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

1 Answer 1


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.

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .