Infinite union of closed sets is not closed

[Mr Davis 97]

Homework Statement

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

Homework Equations

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?

 

[fresh_42]

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

 

[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 $1$?

 

[fresh_42]

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.

 

[LCKurtz]

Just a suggestion to shorten the work, your counterexample doesn't have to be open on both ends. You could use $[\frac 1 n,1]$.

 

[haruspex]

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.

 

[pasmith]

There's always $$\bigcup_{x \in \mathbb{Q}} \{x\} = \mathbb{Q}.$$

 

[WWGD]

Homework Statement

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

Homework Equations

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.

 

[pasmith]

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.

 

[WWGD]

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.

 

[WWGD]

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.

 

[pasmith]

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.

 

[WWGD]

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]
But the thread is about union of closed sets being closed. The name of the thread does not mention the Reals. But, ok, you were only considering Rationals, that is reasonable. But the problem remains . You stated , in regards to my post , that the reals are open and closed and therefore closed. If this is true, then the same is the case for the Rationals, as a space, which is then open and closed and , per your statement, closed. Then the Rationals ate a closed metric space. But then they must be complete. But they ate not. This contradiction follows from the assumption that a whole space being open and closed must then be closed.

 

[StoneTemplePython]

But the thread is about union of closed sets being closed. The name of the thread does not mention the Reals.

I am pretty sure this is from a real analysis course and hence $\mathbb R$ is the overwhelming emphasis. On several occasions recently I've had the thought that the HW template should indicate what the course (or text?) is so we know what OP is oriented toward

In a sense, bring the whoole space, it must be closed, but this may be open ( ha-ha) to interpretation.

I liked this quite a bit though