Heine-Borel in $\mathbb{R}^2$

Question

I want to prove this variation of Heine-Borel theorem in $\mathbb{R}^2$ in the following way.

Theorem. The square $C=[c_1-r,c_1+r]\times[c_2-r,c_2+r]$ is compact.

Here is the idea for the proof.

Let $\mathfrak{F}$ be an open cover made of square balls (With the max metric) for $C$ (This means the elements of $\mathfrak{F}$ are of the form $I\times J$ for $I,J$ open intervals of the same lenght). Let $x\in[c_1-r,c_1+r],y\in[c_2-r,c_2+r]$.

Then $(x,y)\in C$ and there exists some intervals $I,J$ such that $I\times J\in \mathfrak{F}$ and $x\in I$, and $y\in J$. This means that the family $$\mathfrak{F}_X=\{I\mid I\times J\in\mathfrak{F},J\text{ any interval}\}$$ is an open cover for $[c_1-r,c_1+r]$. Then by Heine-Borel in $\mathbb{R}$ there exists a finite subcover of $\mathfrak{F}_X$, let's call it $\mathfrak{F}_X'$. Analogously we get a finite subcover $\mathfrak{F}_Y'$ of $$\mathfrak{F}_Y=\{J\mid I\times J\in\mathfrak{F},I\text{ any interval}\}$$

Using these subcovers I want to build a finite subcover of $\mathfrak{F}$. I wanted to build the family: $\mathfrak{F}'=\{I\times J \mid I\in\mathfrak{F}_X', J\in \mathfrak{F}_Y'\}$, but I cannot assure that these elements $I\times J$ are in $\mathfrak{F}$. How can I finish this?

Answer 1

It seems to me that those elements are really not in $\mathfrak{F}$. Instead, we can use the following method (Essentially we are using Henie Borel theorem twice):

First fix $x\in [c_1 - r, c_1+r]$. Then for each $(x, y)$, there is an open set $U_{(x, y)} = I_{(x, y)} \times J_{(x, y)}$ so that $(x, y) \in U_{(x, y)}$. In particular, $\{J_{(x, y)}\}_{y\in[c_2-r, c_2+r]}$ is an open cover of $[c_2-r, c_2 + r]$ and by Heine Borel theorem there is $k$ (might depends on $x$) so that $[c_2 -r, c_2 + r]$ is covered by

$$J_{(x, y_1)}, J_{(x, y_2)}, \cdots, J_{(x, y_k)} . $$

Let $$I_x = \bigcap_{i=1}^k I_{(x, y_i)}.$$

This $I_x$ is open and $x\in I_x$. Thus $\{I_x\}_{x\in [c_1-r, c_1+r]}$ is an open cover of $[c_1-r, c_1 + r]$ and by Heine Borel theorem again, there is $x_1, \cdots, x_m$ so that $[c_1-r, c_1 +r]$ is covered by $I_{x_1}, I_{x_2}, \cdots, I_{x_m}$. Then the subset of $\mathfrak F$ given by

$$ \left\{ U_{(x_i, y_j)} \right\}_{i=1, \cdots, m, j = 1, \cdots, k_i}$$

is finite and is an open cover of $[c_1-r, c_1 + r] \times [c_2 - r, c_2 + r]$. To see this, let $(x, y)$ be an element in this set. Then there is $i$ so that $x\in I_{x_i}$ and $j$ so that $y\in J_{(x_i, y_j)}$. Now as

$$I_{x_i } = \bigcup_{j=1}^{k_j} I_{(x_i, y_j)} \subset I_{(x_i, y_j)},$$

we have $$(x, y) \in I_{x_i} \times J_{(x_i, y_j)} \subset I_{(x_i, y_j)} \times J_{(x_i, y_j)} = U_{(x_i, y_j)}.$$

Thus every open cover $\mathfrak F$ has a finite subcover.

We remark that

• The proof can be somehow easier if you use sequential compactness (which is eqviualent to Heine Borel Property).

• The above proof actually generalize a lot: It shows that $X\times Y$ is compact if $X, Y$ are both compact.

More information can be found in this question. Indeed, my answer here is just a one of the answer there, with small modifications to follow your notations)