MHB The rectangle has an empty interior

[mathmari]

Hey!

Show that the measure of a rectangle is zero if and only if it has an empty interior.

When a rectangle has an empty interior, does this mean that the length of the sides of the rectangle are equal to zero?? (Wondering)

 

[Euge]

Hey!

Show that the measure of a rectangle is zero if and only if it has an empty interior.

When a rectangle has an empty interior, does this mean that the length of the sides of the rectangle are equal to zero?? (Wondering)

No, it means that the rectangle has no interior points.

 

[mathmari]

No, it means that the rectangle has no interior points.

Ahaa... Ok..

But how can that be, when the measure is equal to the volume=area of the rectangle, which is not equal to zero when the rectangle has no interior points?? (Wondering)

 

[Euge]

Ahaa... Ok..

But how can that be, when the measure is equal to the volume=area of the rectangle, which is not equal to zero when the rectangle has no interior points?? (Wondering)

Let's consider the 1-dimensional case. A rectangle $E$ with no interior points must be either empty or degenerate. In both cases, its measure is zero.

Conversely, suppose $E$ has nonempty interior. Then $E$ contains an interval of the form $(a,b)$. Thus $m(E) \ge b - a > 0$.

 

[mathmari]

Let's consider the 1-dimensional case. A rectangle $E$ with no interior points must be either empty or nondegenerate. In both cases, its measure is zero.

Conversely, suppose $E$ has nonempty interior. Then $E$ contains an interval of the form $(a,b)$. Thus $m(E) \ge b - a > 0$.

At the 1-dimensional case the rectangle is an interval, right??

An interval with no interior points, must be empty. So can we just say that the measure is zero, or do we have to prove it??

Could you explain me what it means that it is nondegenerate?? (Worried)

When we suppose that E has nonempty interior, why is it $m(E) \geq b-a$ and not $m(E)=b-a$ ??

The above is at the one dimension. How could we say it generally?? Do we have to prove this by induction?? (Wondering)

 

[Euge]

At the 1-dimensional case the rectangle is an interval, right??

An interval with no interior points, must be empty. So can we just say that the measure is zero, or do we have to prove it??

Could you explain me what it means that it is nondegenerate?? (Worried)

When we suppose that E has nonempty interior, why is it $m(E) \geq b-a$ and not $m(E)=b-a$ ??

The above is at the one dimension. How could we say it generally?? Do we have to prove this by induction?? (Wondering)

Yes, a 1-D rectangle is an interval. Sorry, I meant to say that an interval with no interior points is either empty or degenerate (i.e., a one-point set). Since $E\supset (a,b)$, $m(E) \ge m((a,b)) = b - a$. You can't deduce $m(E) = b - a$ from $E \supset (a,b)$.

For the general case, start by showing that a rectangle with nonempty interior has positive measure. There's no need for induction.

 

[mathmari]

Yes, a 1-D rectangle is an interval. Sorry, I meant to say that an interval with no interior points is either empty or degenerate (i.e., a one-point set). Since $E\supset (a,b)$, $m(E) \ge m((a,b)) = b - a$. You can't deduce $m(E) = b - a$ from $E \supset (a,b)$.

Ahaa... Ok! (Smile)

For the general case, start by showing that a rectangle with nonempty interior has positive measure. There's no need for induction.

Do we show this by using the fact that the measure of the rectangle is equal to the volume of the rectangle which is not equal to zero, since it has an nonempty interior?? (Wondering)

 

[Euge]

Do we show this by using the fact that the measure of the rectangle is equal to the volume of the rectangle which is not equal to zero, since it has an nonempty interior?? (Wondering)

Not quite. Use the fact that a rectangle $E$ of nonempty interior contains an open rectangle, say $(a_1, b_1) \times \cdots \times (a_n, b_n)$.

 

[mathmari]

Not quite. Use the fact that a rectangle $E$ of nonempty interior contains an open rectangle, say $(a_1, b_1) \times \cdots \times (a_n, b_n)$.

I got stuck right now... (Worried)

Could you explain it further to me?? (Wondering)

 

[Euge]

With the above fact, you deduce

$$m(E) \ge \text{vol}[(a_1, b_1) \times (a_n, b_n)] = \prod_{j = 1}^n (b_j - a_j) > 0,$$

since $b_j > a_j$ for all $j$.

 

[mathmari]

With the above fact, you deduce

$$m(E) \ge \text{vol}[(a_1, b_1) \times (a_n, b_n)] = \prod_{j = 1}^n (b_j - a_j) > 0,$$

since $b_j > a_j$ for all $j$.

Ahaa... Ok! I got it now..So, is it as followed??

$\Rightarrow :$

The measure of a rectangle $E$ is zero.
We suppose that it has a nonempty interior, so it contains an open rectangle $(a_1, b_1) \times \dots \times (a_n, b_n)$.
Then $m(E) \geq v \left ( (a_1, b_1) \times \dots \times (a_n, b_n) \right )=\prod_{j = 1}^n (b_j - a_j) > 0$.
That cannot be true, since $m(E)=0$.
Therefore, if the measure of a rectangle is zero then it has an empty interior.

$\Leftarrow :$

The rectangle has an empty interior.
The rectangle is the union of the interior of the rectangle ($I$) and the bounds of the rectangle ($B$).
So, $m(E)=m(I \cup B)=m(I)+m(B)$.
Is this correct?? How could I continue?? (Wondering)

 

[Euge]

Ahaa... Ok! I got it now..So, is it as followed??

$\Rightarrow :$

The measure of a rectangle $E$ is zero.
We suppose that it has a nonempty interior, so it contains an open rectangle $(a_1, b_1) \times \dots \times (a_n, b_n)$.
Then $m(E) \geq v \left ( (a_1, b_1) \times \dots \times (a_n, b_n) \right )=\prod_{j = 1}^n (b_j - a_j) > 0$.
That cannot be true, since $m(E)=0$.
Therefore, if the measure of a rectangle is zero then it has an empty interior.

$\Leftarrow :$

The rectangle has an empty interior.
The rectangle is the union of the interior of the rectangle ($I$) and the bounds of the rectangle ($B$).
So, $m(E)=m(I \cup B)=m(I)+m(B)$.
Is this correct?? How could I continue?? (Wondering)

The first part looks good. The second part needs work. For the reverse direction ($\Leftarrow$), let $E$ be your rectangle, and write it as $E = \{(x_1,x_2\ldots, x_n)\, |\, a_j \le x_j \le b_j,\, j = 1, 2,\ldots, n)\}$ where the inequalities may not all be sharp. If $E$ has empty interior, then there is a $j$ for which $b_j = a_j$ (in other words, $E$ is degenerate). Otherwise $E$ contains the open rectangle $(a_1,b_1) \times \cdots (a_n,b_n)$, which contradicts the fact that $E$ contains no nonempty open set.

If $a_j = b_j$ for all $j$, then $E = \{(a_1,a_2\ldots, a_n)\}$ and $m(E) = 0$. Otherwise, suppose $a_j = b_j$ for only $r$ values of $j$, where $r < n$. Without loss of generality, assume $a_1 = b_1, a_2 = b_2, \ldots, a_r = b_r$. Given $\varepsilon > 0$, $E$ is covered by open rectangles of the form

$$R_j :=
\bigl(a_1 - \tfrac{\varepsilon}{2^{j+1}}, a_1 + \tfrac{\varepsilon}{2^{j+1}}\bigr) \times \cdots \times \bigl(a_r - \tfrac{\varepsilon}{2^{j+1}}, a_r + \tfrac{\varepsilon}{2^{j+1}}\bigr) \times (a_{r+1}, b_{r+1}) \times \cdots \times (a_n, b_n), \quad j = 1, 2, \ldots$$

and

$$\sum_j v(R_j) = \sum_j \frac{r\varepsilon}{2^j}\prod_{r < i \le n}(b_j - a_j) = C\varepsilon,\quad C = r\prod_{r < i \le n} (b_j - a_j).$$

Therefore $m(E) \le C\varepsilon$. Since $\varepsilon$ was arbitrary, we deduce $m(E) = 0$.

 

[mathmari]

For the reverse direction ($\Leftarrow$), let $E$ be your rectangle, and write it as $E = \{(x_1,x_2\ldots, x_n)\, |\, a_j \le x_j \le b_j,\, j = 1, 2,\ldots, n)\}$ where the inequalities may not all be sharp. If $E$ has empty interior, then there is a $j$ for which $b_j = a_j$ (in other words, $E$ is degenerate). Otherwise $E$ contains the open rectangle $(a_1,b_1) \times \cdots (a_n,b_n)$, which contradicts the fact that $E$ contains no nonempty open set.

If $a_j = b_j$ for all $j$, then $E = \{(a_1,a_2\ldots, a_n)\}$ and $m(E) = 0$. Otherwise, suppose $a_j = b_j$ for only $r$ values of $j$, where $r < n$. Without loss of generality, assume $a_1 = b_1, a_2 = b_2, \ldots, a_r = b_r$. Given $\varepsilon > 0$, $E$ is covered by open rectangles of the form

$$R_j :=
\bigl(a_1 - \tfrac{\varepsilon}{2^{j+1}}, a_1 + \tfrac{\varepsilon}{2^{j+1}}\bigr) \times \cdots \times \bigl(a_r - \tfrac{\varepsilon}{2^{j+1}}, a_r + \tfrac{\varepsilon}{2^{j+1}}\bigr) \times (a_{r+1}, b_{r+1}) \times \cdots \times (a_n, b_n), \quad j = 1, 2, \ldots$$

and

$$\sum_j v(R_j) = \sum_j \frac{r\varepsilon}{2^j}\prod_{r < i \le n}(b_j - a_j) = C\varepsilon,\quad C = r\prod_{r < i \le n} (b_j - a_j).$$

Therefore $m(E) \le C\varepsilon$. Since $\varepsilon$ was arbitrary, we deduce $m(E) = 0$.

So, when the rectangle $E = \{(x_1,x_2\ldots, x_n)\, |\, a_j \le x_j \le b_j,\, j = 1, 2,\ldots, n)\}$ has an empty interior, that means that $a_j = b_j$ for all $j$, or for $r$ values of $j$, where $r < n$ ?? (Wondering)

Could you explain me why this stands?? (Worried)

 

[Euge]

So, when the rectangle $E = \{(x_1,x_2\ldots, x_n)\, |\, a_j \le x_j \le b_j,\, j = 1, 2,\ldots, n)\}$ has an empty interior, that means that $a_j = b_j$ for all $j$, or for $r$ values of $j$, where $r < n$ ?? (Wondering)

Could you explain me why it stands?? (Worried)

I showed that when $E$ has empty interior, $a_j = b_j$ for at least one $j$. So I considered two cases. In the first case, $a_j = b_j$ for all $j$. In the second case, $a_j$ is not equal to $b_j$ for all $j$. Therefore, there is an $r < n$ such that $a_j = b_j$ for only $r$ values of $j$ (so the other $n - r$ values of $j$ have $a_j \neq b_j$).

 

[mathmari]

I showed that when $E$ has empty interior, $a_j = b_j$ for at least one $j$. So I considered two cases. In the first case, $a_j = b_j$ for all $j$. In the second case, $a_j$ is not equal to $b_j$ for all $j$. Therefore, there is an $r < n$ such that $a_j = b_j$ for only $r$ values of $j$ (so the other $n - r$ values of $j$ have $a_j \neq b_j$).

Ahaa... Ok! I see... (Smile)

For the reverse direction ($\Leftarrow$), let $E$ be your rectangle, and write it as $E = \{(x_1,x_2\ldots, x_n)\, |\, a_j \le x_j \le b_j,\, j = 1, 2,\ldots, n)\}$ where the inequalities may not all be sharp. If $E$ has empty interior, then there is a $j$ for which $b_j = a_j$ (in other words, $E$ is degenerate). Otherwise $E$ contains the open rectangle $(a_1,b_1) \times \cdots (a_n,b_n)$, which contradicts the fact that $E$ contains no nonempty open set.

When we write the rectangle in the form $E = \{(x_1,x_2\ldots, x_n)\, |\, a_j \le x_j \le b_j,\, j = 1, 2,\ldots, n)\}$, are the $x_j$ intervals??

So, is $E$ of the form $$E=[a_1,b_1] \times [a_2,b_2] \times \dots \times [a_n,b_n] \text{ or } E=(a_1,b_1) \times (a_2,b_2) \times \dots \times (a_n,b_n)$$ ?? (Wondering)

If $a_j = b_j$ for all $j$, then $E = \{(a_1,a_2\ldots, a_n)\}$ and $m(E) = 0$.

Do we have that $m(E) = 0$, because $$m(E) =m \left ( \cup_{i=0}^{n} (a_i,a_i) \right )=\sum_{i=0}^{n} m((a_i,a_i))=\sum_{i=0}^{n} (a_i-a_i)=0$$ ?? (Wondering)

Otherwise, suppose $a_j = b_j$ for only $r$ values of $j$, where $r < n$. Without loss of generality, assume $a_1 = b_1, a_2 = b_2, \ldots, a_r = b_r$. Given $\varepsilon > 0$, $E$ is covered by open rectangles of the form

$$R_j :=
\bigl(a_1 - \tfrac{\varepsilon}{2^{j+1}}, a_1 + \tfrac{\varepsilon}{2^{j+1}}\bigr) \times \cdots \times \bigl(a_r - \tfrac{\varepsilon}{2^{j+1}}, a_r + \tfrac{\varepsilon}{2^{j+1}}\bigr) \times (a_{r+1}, b_{r+1}) \times \cdots \times (a_n, b_n), \quad j = 1, 2, \ldots$$

and

$$\sum_j v(R_j) = \sum_j \frac{r\varepsilon}{2^j}\prod_{r < i \le n}(b_j - a_j) = C\varepsilon,\quad C = r\prod_{r < i \le n} (b_j - a_j).$$

Therefore $m(E) \le C\varepsilon$. Since $\varepsilon$ was arbitrary, we deduce $m(E) = 0$.

Is $E$ covered by open rectangles only if it is of the form $E=(a_1,b_1) \times (a_2,b_2) \times \dots \times (a_n,b_n)$ ??

When it were of the form $E=[a_1,b_1] \times [a_2,b_2] \times \dots \times [a_n,b_n]$, do we have to cover $E$ by closed rectangles?? (Wondering)

 

[Euge]

When we write the rectangle in the form $E = \{(x_1,x_2\ldots, x_n)\, |\, a_j \le x_j \le b_j,\, j = 1, 2,\ldots, n)\}$, are the $x_j$ intervals??

So, is $E$ of the form $$E=[a_1,b_1] \times [a_2,b_2] \times \dots \times [a_n,b_n] \text{ or } E=(a_1,b_1) \times (a_2,b_2) \times \dots \times (a_n,b_n)$$ ?? (Wondering)

No to both questions. The points $(x_1,x_2,\ldots, x_n)$ are elements of $\Bbb R^n$. Also, intervals are sets, not real numbers. So conceptually speaking the $x_j$ cannot be intervals. Since the inequalities $a_j \le x_j \le b_j$ may not all be sharp, $E$ may be neither open nor closed. So it may not be any of the two forms you've written.

Do we have that $m(E) = 0$, because $$m(E) =m \left ( \cup_{i=0}^{n} (a_i,a_i) \right )=\sum_{i=0}^{n} m((a_i,a_i))=\sum_{i=0}^{n} (a_i-a_i)=0$$ ?? (Wondering)

No. The expressions $m(a_i, a_i)$ do not make sense. We have $m(E) = 0$ because for every $\varepsilon > 0$, $E$ is covered by open rectangles of the form

$$R_j = \prod_{i = 1}^n \Bigl(a_i - \frac{\varepsilon}{2^{j+2}}, a_i + \frac{\varepsilon}{2^{j+2}}\Bigr), \quad j = 1, 2, \ldots$$

with

$$ \sum_j v(R_j) = \sum_j \frac{\varepsilon}{2^{j+1}} = \frac{\varepsilon}{2} < \varepsilon.$$

Is $E$ covered by open rectangles only if it is of the form $E=(a_1,b_1) \times (a_2,b_2) \times \dots \times (a_n,b_n)$ ??

When it were of the form $E=[a_1,b_1] \times [a_2,b_2] \times \dots \times [a_n,b_n]$, do we have to cover $E$ by closed rectangles?? (Wondering)

The answer to the first question is no. Certainly the rectangle $R = [0,1) \times \cdots \times [0, 1)$ is covered by the closed rectangle $[0,1] \times \cdots \times [0,1]$ even though $R$ is not closed.

I don't understand where you're going with the second question. When was $E$ of that form?