# Infinite union of closed sets is not closed

• Mr Davis 97
Mr Davis 97

## Homework Statement

Show that it is not necessarily true that the infinite union of closed sets is closed

## The Attempt at a Solution

From intuition, I came up with the following counter-example: ##\displaystyle \bigcup_{n=2}^{\infty} \left[ \frac{1}{n}, \frac{n}{n+1} \right] = (0,1)##. However, I am not sure how to actually prove that this infintie union is the set on the right. How would I do so?

Mentor
2022 Award
As always with sets. You prove ##\bigcup_n [\frac{1}{n}, \frac{n}{n+1}] \subseteq (0,1)## and ##(0,1) \subseteq \bigcup_n [\frac{1}{n}, \frac{n}{n+1}]\,.##

member 587159
Mr Davis 97
As always with sets. You prove ##\bigcup_n [\frac{1}{n}, \frac{n}{n+1}] \subseteq (0,1)## and ##(0,1) \subseteq \bigcup_n [\frac{1}{n}, \frac{n}{n+1}]\,.##
So if I let ##x\in \displaystyle \bigcup_{n=2}^{\infty} \left[ \frac{1}{n}, \frac{n}{n+1} \right]##, I want to show that ##x\in (0,1)##. However, I don't see what ##x## being in ##\displaystyle \bigcup_{n=2}^{\infty} \left[ \frac{1}{n}, \frac{n}{n+1} \right]## really tells me. Do we just have to note that if ##x## is in this union then it is always greater than ##0## and less than ##1##?

Mentor
2022 Award
So if I let ##x\in \displaystyle \bigcup_{n=2}^{\infty} \left[ \frac{1}{n}, \frac{n}{n+1} \right]##, I want to show that ##x\in (0,1)##. However, I don't see what ##x## being in ##\displaystyle \bigcup_{n=2}^{\infty} \left[ \frac{1}{n}, \frac{n}{n+1} \right]## really tells me. Do we just have to note that if ##x## is in this union then it is always greater than ##0## and less than ##1##?
No and yes. If you don't know what it tells you, then you should insert a step in between. However, in the end, yes.

If an element ##x## is in a union of sets, then it has to be in at least one of those sets. In our case this means ##x \in [\frac{1}{n},\frac{n}{n+1}]## for some ##n>1##, i.e. ##0 < \frac{1}{n} \leq x \leq \frac{n}{n+1} < 1## which is the step in between.

And now for the other direction.

Mr Davis 97
Homework Helper
Gold Member
Just a suggestion to shorten the work, your counterexample doesn't have to be open on both ends. You could use ##[\frac 1 n,1]##.

Homework Helper
Gold Member
2022 Award
not sure how to actually prove that this infintie union is the set on the right.
You do not need to. You just need to find a convergent sequence of points with a limit outside the set.

Homework Helper
2022 Award
There's always $$\bigcup_{x \in \mathbb{Q}} \{x\} = \mathbb{Q}.$$

WWGD
Gold Member

## Homework Statement

Show that it is not necessarily true that the infinite union of closed sets is closed

## The Attempt at a Solution

From intuition, I came up with the following counter-example: ##\displaystyle \bigcup_{n=2}^{\infty} \left[ \frac{1}{n}, \frac{n}{n+1} \right] = (0,1)##. However, I am not sure how to actually prove that this infintie union is the set on the right. How would I do so?
Or, more drastically, do the union of all r in Reals. It is closed and open, since it is the whole space.

Homework Helper
2022 Award
Or, more drastically, do the union of all r in Reals. It is closed and open, since it is the whole space.

But it is closed, so it's a union of closed sets which is closed, but happens also to be open.

fresh_42
Gold Member
But it is closed, so it's a union of closed sets which is closed, but happens also to be open.
LL, this has always been kind of confusing . Is a set that is both closed and open strictly closed? In a sense, bring the whoole space, it must be closed, but this may be open ( ha-ha) to interpretation. Notice it is complete as a closed metric space should be, but the same is not the case for the Rationals. So, to be less ambiguous, I propose(edit ) the union of alln singleton Rationals as Real numbers.

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StoneTemplePython
Gold Member
There's always $$\bigcup_{x \in \mathbb{Q}} \{x\} = \mathbb{Q}.$$
But, you see, if you consider ##\mathbb Q## as a stand alone space, then it is open and closed and, according to your statement, closed. But this is problematic if you believe this is equivalent to being closed because a closed metric space must be complete, and the Rationals are not.

Homework Helper
2022 Award
But, you see, if you consider ##\mathbb Q## as a stand alone space,

The entire thread is dealing with the real numbers in the usual topology, so $\mathbb{Q} \subsetneq \mathbb{R}$ is neither open nor closed.

Butthethreah, post: 6103787, member: 415692"]The entire thread is dealing with the real numbers in the usual topology, so $\mathbb{Q} \subsetneq \mathbb{R}$ is neither open nor closed.[/QUOTE]