MHB An open cover is a collection of open sets whose union contains a given subset

[mathmari]

Hey!

Let $\mathbb{R}$ provided with the metric $d(x,y)=|x-y|$. I want to check if the collections of sets $$S_1=\left \{\left (\frac{x}{2}, \frac{3x}{2}\right ): 0<x<1\right \}, \ \ \ \ \ S_2=\left \{\left (x-\frac{1}{2}, x+\frac{1}{2}\right ): 0<x<1\right \}$$ are open covers of $A=\left \{\frac{1}{n} : n\in \mathbb{N}\right \}\subset \mathbb{R}$.

An open cover is a collection of open sets whose union contains a given subset, right?

Could you give me a hint how we could check that in these cases? Do we have to check if the union of $S_1$ of all $x$ contains $A$ and the same also for $S_2$ ? (Wondering)

 

[S.G. Janssens]

Hey!

Let $\mathbb{R}$ provided with the metric $d(x,y)=|x-y|$. I want to check if the collections of sets $$S_1=\left \{\left (\frac{x}{2}, \frac{3x}{2}\right ): 0<x<1\right \}, \ \ \ \ \ S_2=\left \{\left (x-\frac{1}{2}, x+\frac{1}{2}\right ): 0<x<1\right \}$$ are open covers of $A=\left \{\frac{1}{n} : n\in \mathbb{N}\right \}\subset \mathbb{R}$.

An open cover is a collection of open sets whose union contains a given subset, right?

That is right.

Could you give me a hint how we could check that in these cases? Do we have to check if the union of $S_1$ of all $x$ contains $A$ and the same also for $S_2$ ? (Wondering)

Yes. In some more detail: You need to check if for every $a \in A$ there exists an interval $I$ in the collection $S_1$ (generally, $I$ will depend on $a$) such that $a \in I$, and similarly for $S_2$.

Since such $a \in A$ is of the form $a = \frac{1}{n}$ for some $n \in \mathbb{N}$, and each interval $I$ in $S_1$ is of the form $I = \left(\frac{x}{2},\frac{3x}{2}\right)$ for some $0 < x < 1$, what you really need to check is: Is there for every $n \in \mathbb{N}$ an $0 < x < 1$ such that $\frac{1}{n} \in \left(\frac{x}{2},\frac{3x}{2}\right)$?

 

[mathmari]

Since such $a \in A$ is of the form $a = \frac{1}{n}$ for some $n \in \mathbb{N}$, and each interval $I$ in $S_1$ is of the form $I = \left(\frac{x}{2},\frac{3x}{2}\right)$ for some $0 < x < 1$, what you really need to check is: Is there for every $n \in \mathbb{N}$ an $0 < x < 1$ such that $\frac{1}{n} \in \left(\frac{x}{2},\frac{3x}{2}\right)$?

I think that it holds that there is for every $n \in \mathbb{N}$ an $0 < x < 1$ such that $\frac{1}{n} \in \left(\frac{x}{2},\frac{3x}{2}\right)$, because the maximum value of $\frac{1}{n}$ is $1$ which is in the interval $\left(\frac{x}{2},\frac{3x}{2}\right)$ for $x$ that tends to $1$ and the minimum value tends to $0$ which is also in the interval $\left(\frac{x}{2},\frac{3x}{2}\right)$ if $x$ tends to $0$, or not?

If this is correct, how could we prove that formally? (Wondering)

 

[S.G. Janssens]

I think that it holds that there is for every $n \in \mathbb{N}$ an $0 < x < 1$ such that $\frac{1}{n} \in \left(\frac{x}{2},\frac{3x}{2}\right)$, because the maximum value of $\frac{1}{n}$ is $1$ which is in the interval $\left(\frac{x}{2},\frac{3x}{2}\right)$ for $x$ that tends to $1$ and the minimum value tends to $0$ which is also in the interval $\left(\frac{x}{2},\frac{3x}{2}\right)$ if $x$ tends to $0$, or not?

If this is correct, how could we prove that formally? (Wondering)

Yes, you have the right intuition.

Now you need to show that for any given $n \in \mathbb{N}$ the two equalities
\[
\frac{x}{2} < \frac{1}{n} < \frac{3 x}{2}
\]
have at least one solution $x \in (0,1)$. This is a matter of rewriting: Multiply by $2$ to obtain the equivalent system
\[
x < \frac{2}{n} < 3x.
\]
Can you now see a very simple expression for $x$ in terms of $n$ that satisfies this system and is such that $x \in (0,1)$?

 

[mathmari]

\[
x < \frac{2}{n} < 3x.
\]
Can you now see a very simple expression for $x$ in terms of $n$ that satisfies this system and is such that $x \in (0,1)$?

We have that $x<\frac{2}{n}<1$ for $n>2$ and $3x>\frac{2}{n}\Rightarrow x>\frac{2}{3n}>0$ for all $n$.

Is this enough to conclude that $x\in (0,1)$ ? (Wondering)

 

[S.G. Janssens]

We have that $x<\frac{2}{n}<1$ for $n>2$ and $3x>\frac{2}{n}\Rightarrow x>\frac{2}{3n}>0$ for all $n$.

Is this enough to conclude that $x\in (0,1)$ ? (Wondering)

It is enough to conclude that, if there exists a real number $x$ depending on $n$ that satisfies
\[
x < \frac{2}{n} < 3x \qquad (*)
\]
then $0 < x < 1$.

However, this is not enough to finish the question. Namely, what you need to show is that for each $n \in \mathbb{N}$ there exists some $x \in (0,1)$ - depending on $n$ - such that (*) holds.

Can you see how this is different from what you wrote?

Can you find a simple formula for $x$ in terms of $n$ that accomplishes the latter? (Do not think too hard about it: For what expressions for $x$ involving $n$ is (*) satisfied?)

 

[mathmari]

Can you find a simple formula for $x$ in terms of $n$ that accomplishes the latter? (Do not think too hard about it: For what expressions for $x$ involving $n$ is (*) satisfied?)

Is it maybe $x=\frac{1}{n}$ ? Then (*) would be satisfied. Or I am thinking wrong? (Wondering)

 

[S.G. Janssens]

Is it maybe $x=\frac{1}{n}$ ? Then (*) would be satisfied. Or I am thinking wrong? (Wondering)

Right! (Yes)

 

[mathmari]

Right! (Yes)

But for $n=1$ we have that $x=1$ and then $x\notin (0,1)$, or not? (Wondering)
As for $S_2$:

$\displaystyle{ S_2=\left \{\left (x-\frac{1}{2}, x+\frac{1}{2}\right ): 0<x<1\right \}}$

We have to check if for every $a \in A$ there exists an interval $I$ in the collection $S_2$ such that $a \in I$.
Since such $a \in A$ is of the form $a = \frac{1}{n}$ for some $n \in \mathbb{N}$, and each interval $I$ in $S_2$ is of the form $I = \left (x-\frac{1}{2}, x+\frac{1}{2}\right )$ for some $0 < x < 1$, we want to check the following:
Is there for every $n \in \mathbb{N}$ an $0 < x < 1$ such that $\frac{1}{n} \in \left (x-\frac{1}{2}, x+\frac{1}{2}\right )$?
For each $n \in \mathbb{N}$ the inequalities $x-\frac{1}{2} < \frac{1}{n} < x+\frac{1}{2}$ must have at least one solution $x \in (0,1)$.

Do we have to find here again an expression of $x$ in terms of $n$ ? We have that $\left |x-\frac{1}{n}\right |<\frac{1}{2}$ ans also $\frac{1}{n}-\frac{1}{2}<x<\frac{1}{n}+\frac{1}{2}$, right? But how can we get an expression? (Wondering)

 

[S.G. Janssens]

But for $n=1$ we have that $x=1$ and then $x\notin (0,1)$, or not? (Wondering)

You are certainly right. I should have added that the initial case $n = 1$ is covered (pun intended) by e.g. $x = \frac{3}{4}$. So, in summary,

Yes, $S_1$ is an open cover of $A$, because for every point $a \in A$ the interval $\left(\frac{x}{2},\frac{3x}{2} \right) \in S_1$ with
\[
x =
\begin{cases}
a & \text{if } a < 1,\\
\frac{3}{4} & \text{if } a = 1,
\end{cases}
\]
is such that $a \in S_1$.

As for $S_2$:

$\displaystyle{ S_2=\left \{\left (x-\frac{1}{2}, x+\frac{1}{2}\right ): 0<x<1\right \}}$

We have to check if for every $a \in A$ there exists an interval $I$ in the collection $S_2$ such that $a \in I$.
Since such $a \in A$ is of the form $a = \frac{1}{n}$ for some $n \in \mathbb{N}$, and each interval $I$ in $S_2$ is of the form $I = \left (x-\frac{1}{2}, x+\frac{1}{2}\right )$ for some $0 < x < 1$, we want to check the following:
Is there for every $n \in \mathbb{N}$ an $0 < x < 1$ such that $\frac{1}{n} \in \left (x-\frac{1}{2}, x+\frac{1}{2}\right )$?
For each $n \in \mathbb{N}$ the inequalities $x-\frac{1}{2} < \frac{1}{n} < x+\frac{1}{2}$ must have at least one solution $x \in (0,1)$.

Yes, very good.

Do we have to find here again an expression of $x$ in terms of $n$ ? We have that $\left |x-\frac{1}{n}\right |<\frac{1}{2}$ ans also $\frac{1}{n}-\frac{1}{2}<x<\frac{1}{n}+\frac{1}{2}$, right?

That depends on what you choose for $x$. For example, if $n = 10$ and $x = \frac{9}{10}$ then neither is true.
Since your statement does not have quantifiers (in words or symbols), it is not unambiguous.

But how can we get an expression? (Wondering)

Begin by covering for special cases. For example, for $n = 1$ we can again choose $x = \frac{3}{4}$. Then consider some simple expressions in terms of $n$ that take values in $(0,1)$ - the interval in which $x$ should lie. You can use a familiar result.

 

[S.G. Janssens]

Incidentally, for $S_2$ there is also another approach possible.

You can first deal with the case $n = 1$ as before. Next, you can find one $ x \in (0,1)$ - not depending on $n$ - such that $\left(x - \frac{1}{2}, x + \frac{1}{2}\right)$ contains $\frac{1}{n}$ for all $n > 1$. There are many possibilities to choose for $x$ in this way. Can you find some?

Incidentally, this is somewhat rare. In general, the open set containing the point will indeed depend on the point in question.

 

[mathmari]

You are certainly right. I should have added that the initial case $n = 1$ is covered (pun intended) by e.g. $x = \frac{3}{4}$. So, in summary,

Yes, $S_1$ is an open cover of $A$, because for every point $a \in A$ the interval $\left(\frac{x}{2},\frac{3x}{2} \right) \in S_1$ with
\[
x =
\begin{cases}
a & \text{if } a < 1,\\
\frac{3}{4} & \text{if } a = 1,
\end{cases}
\]
is such that $a \in S_1$.

Ahh ok, I got it! (Smile)
For $S_2$, do we have the following?

$S_2$ is an open cover of $A$, because for every point $a \in A$ the interval $\left(x-\frac{1}{2},x+\frac{1}{2} \right) \in S_2$ with
$$x =
\begin{cases}
a & \text{if } a < 1,\\
\frac{3}{4} & \text{if } a = 1,
\end{cases}
\ \ \ \ \ \text{ or } \ \ \ \ \ x =
\begin{cases}
\frac{1}{2} & \text{if } a < 1,\\
\frac{3}{4} & \text{if } a = 1,
\end{cases}
$$
is such that $a \in S_2$.

Is this correct? (Wondering)

 

[S.G. Janssens]

Ahh ok, I got it! (Smile)
For $S_2$, do we have the following?

$S_2$ is an open cover of $A$, because for every point $a \in A$ the interval $\left(x-\frac{1}{2},x+\frac{1}{2} \right) \in S_2$ with
$$x =
\begin{cases}
a & \text{if } a < 1,\\
\frac{3}{4} & \text{if } a = 1,
\end{cases}
\ \ \ \ \ \text{ or } \ \ \ \ \ x =
\begin{cases}
\frac{1}{2} & \text{if } a < 1,\\
\frac{3}{4} & \text{if } a = 1,
\end{cases}
$$
is such that $a \in S_2$.

Is this correct? (Wondering)

Yes, well done!

 

[mathmari]

Yes, well done!

Great! So both $S_1$ and $S_2$ are open covers of $A$. To check if they have a finite subcover, we have to check if $S_1$ and $S_2$ respectively, have a finite subset which is also a cover of $A$, right?

For example, we have that $\displaystyle{\tilde{S}_1=\left \{\left (\frac{x}{4}, \frac{3x}{4}\right ): 0<x<1\right \}\subseteq S_1}$.

This is not a cover of $A$, since for $n=1$ the element $a$ is not in $\tilde{S}_1$. The same holds also when we consider a subset of the form $\left \{\left (\frac{x}{i}, \frac{3x}{i}\right ): 0<x<1\right \}$ with $i>2$.

Does this mean that $S_1$ has no finite subcover? (Wondering) As for $S_2$ :

Do we consider a subset as for example $\displaystyle{\tilde{S}_2=\left \{\left (x-\frac{1}{4}, x+\frac{1}{2}\right ): 0<x<1\right \}\subseteq S_2}$ ? (Wondering)

 

[S.G. Janssens]

Great! So both $S_1$ and $S_2$ are open covers of $A$. To check if they have a finite subcover, we have to check if $S_1$ and $S_2$ respectively, have a finite subset which is also a cover of $A$, right?

Right.

For example, we have that $\displaystyle{\tilde{S}_1=\left \{\left (\frac{x}{4}, \frac{3x}{4}\right ): 0<x<1\right \}\subseteq S_1}$.

This is not a cover of $A$, since for $n=1$ the element $a$ is not in $\tilde{S}_1$. The same holds also when we consider a subset of the form $\left \{\left (\frac{x}{i}, \frac{3x}{i}\right ): 0<x<1\right \}$ with $i>2$.

Also right.

Does this mean that $S_1$ has no finite subcover? (Wondering)

No. It is indeed the case that $S_1$ has no finite subcover, but you cannot conclude that from what you wrote. Namely, what you wrote only shows that there exist certain subsets of $S_1$ that are not covers of $A$.

When we want to show that no finite subcover exists, it often helps to argue by contradiction. Suppose, therefore, that the open cover $S_1$ of $A$ has a finite subcover $\overline{S}_1$. Let
\[
\overline{x} := \min\left\{x \in (0,1) \,:\, \left(\frac{x}{2}, \frac{3x}{2} \right) \in \overline{S}_1 \right\} > 0.
\]

Why does this minimum exist and why is it positive?

How can you use $\overline{x}$ to show that there exists $n \in \mathbb{N}$ such that $a = \frac{1}{n}$ is not in any element of $\overline{S}_1$?

As for $S_2$ :

Do we consider a subset as for example $\displaystyle{\tilde{S}_2=\left \{\left (x-\frac{1}{4}, x+\frac{1}{2}\right ): 0<x<1\right \}\subseteq S_2}$ ? (Wondering)

Are you sure that $\overline{S}_2$ is a subset of $S_2$?
In fact, given what you already know about $S_2$ from posts #11 and #12, you can answer this question for $S_2$ quite easily.

Question for you: Based on what we discussed so far, can you say whether or not $A$ is compact?

 

[mathmari]

When we want to show that no finite subcover exists, it often helps to argue by contradiction. Suppose, therefore, that the open cover $S_1$ of $A$ has a finite subcover $\overline{S}_1$. Let
\[
\overline{x} := \min\left\{x \in (0,1) \,:\, \left(\frac{x}{2}, \frac{3x}{2} \right) \in \overline{S}_1 \right\} > 0.
\]

Why does this minimum exist and why is it positive?

How can you use $\overline{x}$ to show that there exists $n \in \mathbb{N}$ such that $a = \frac{1}{n}$ is not in any element of $\overline{S}_1$?

We assume that the open cover $S_1$ of $A$ has a non-empty, finite subcover $\overline{S}_1$.

Let \begin{equation*}\overline{x} := \min\left\{x \in (0,1) \,:\, \left(\frac{x}{2}, \frac{3x}{2} \right) \in \overline{S}_1 \right\} > 0\end{equation*} This minimum exists since $(0,1)$ is a non-empty and finite set and it is positiv since $x\in (0,1)$ and so all $x$-values are positiv.

When $n\in \mathbb{N}$ gets very large, then $\frac{1}{n}$ approximates to $0$.
So when $n$ is large enough, then it holds that $\frac{1}{n}\rightarrow 0$ and $0<\frac{1}{n}<\frac{\overline{x}}{2}$. This means that there is a $n\in \mathbb{N}$ such that $\frac{1}{n}$ is less than every element of the collection of systems $\overline{S}_1$. Therefore $\frac{1}{n}$ is not an element of $\overline{S}_1$.

That implies that $\overline{S}_1$ is not a cover of $A$. So $S_1$ has no finite subcover.
Is everything correct? Could I improve something? (Wondering)

Are you sure that $\overline{S}_2$ is a subset of $S_2$?
In fact, given what you already know about $S_2$ from posts #11 and #12, you can answer this question for $S_2$ quite easily.

According post #12, do we consider a set with $x=\frac{1}{2}$ and a set with $x=\frac{3}{4}$, i.e. $\overline{S}_2=\left \{(0,1), \left (\frac{1}{4}, \frac{5}{4}\right )\right \}$ ? Then it holds that $A\subseteq \overline{S}_2$, or not?

(Wondering)

Question for you: Based on what we discussed so far, can you say whether or not $A$ is compact?

We use here the fact that $A$ is compact if each open cover has a finite subcover, right?

So, since the open cover $S_1$ of $A$ has no finite subcover, it follows that $A$ is not compact.

Is this correct? (Wondering)

 

[S.G. Janssens]

We assume that the open cover $S_1$ of $A$ has a non-empty, finite subcover $\overline{S}_1$.

Let \begin{equation*}\overline{x} := \min\left\{x \in (0,1) \,:\, \left(\frac{x}{2}, \frac{3x}{2} \right) \in \overline{S}_1 \right\} > 0\end{equation*} This minimum exists since $(0,1)$ is a non-empty and finite set and it is positiv since $x\in (0,1)$ and so all $x$-values are positiv.

Almost: Of course $(0,1)$ itself is not finite, but the set
\[
\left\{x \in (0,1) \,:\, \left(\frac{x}{2}, \frac{3x}{2} \right) \in \overline{S}_1 \right\}
\]
over which the minimum is taken, is finite, since $\overline{S}_1$ itself is finite.

When $n\in \mathbb{N}$ gets very large, then $\frac{1}{n}$ approximates to $0$.
So when $n$ is large enough, then it holds that $\frac{1}{n}\rightarrow 0$ and $0<\frac{1}{n}<\frac{\overline{x}}{2}$. This means that there is a $n\in \mathbb{N}$ such that $\frac{1}{n}$ is less than every element of the collection of systems $\overline{S}_1$. Therefore $\frac{1}{n}$ is not an element of $\overline{S}_1$.

That implies that $\overline{S}_1$ is not a cover of $A$. So $S_1$ has no finite subcover.

Is everything correct? Could I improve something? (Wondering)

You are very close, but I would improve the formulation. I would write: "(...) such that $\frac{1}{n}$ is less than the left endpoint of any interval in the collection $\overline{S}_1$. Therefore $\frac{1}{n}$ is not contained in any interval (or: element, or: member...) in $\overline{S}_1$."

Note that $S_1$ and its subcover $\overline{S}_1$ have intervals as members, not numbers.

According post #12, do we consider a set with $x=\frac{1}{2}$ and a set with $x=\frac{3}{4}$, i.e. $\overline{S}_2=\left \{(0,1), \left (\frac{1}{4}, \frac{5}{4}\right )\right \}$ ? Then it holds that $A\subseteq \overline{S}_2$, or not?

(Wondering)

Yes, $\overline{S}_2$ as you indicated is a finite subcover, but be careful to note that $A$ consists of numbers while $\overline{S}_2$ consists of intervals, so we cannot write $A \subseteq \overline{S}_2$, but we can say that $A$ is covered by $\overline{S}_2$, since the union of intervals in $\overline{S}_2$ contains $A$.

We use here the fact that $A$ is compact if each open cover has a finite subcover, right?

Right. This is usually taken as the definition of compactness, but that may depend on the choice you book makes.

So, since the open cover $S_1$ of $A$ has no finite subcover, it follows that $A$ is not compact.

Is this correct? (Wondering)

Yes!