Muhammad Haris Rao
wThroughout, $\mathbb{R}^n$ is given the usual topology. We will prove here the following theorem:
Theorem (Heine-Borel): A subset $K \subseteq \mathbb{R}^n$ is compact if and only if it is closed and bounded.
Proposition: Let $(X, d)$ be a metric space. Then all compact subsets are bounded.
Proof. We will show if $K$ is not bounded, then it cannot be compact.
Suppose that $K$ is not bounded. Fix $x_0 \in K$ and for $n \in \mathbb{Z}_{> 0}$, let $U_n$ be the open ball $$ U_n = \{ y \in X \mid d(x_0, y) < n \} $$ Then we have that $\{ U_n \}_{n \in \mathbb{Z}_{> 0}}$ is a cover of $K$ because if $y \in K$ then we can choose $N \in \mathbb{Z}_{> d(x_0, y)}$ so that $y \in U_N \subseteq \bigcup_{n = 1}^\infty U_n$. However, there is not finite subcover. If $C \subseteq \mathbb{Z}_{> 0}$ is finite, Then $$ \bigcup_{n \in C} U_n = U_{m_C} $$ where $m_C = \max \, C$. However, since $K$ is unbouded there exist $y, z \in K$ with $d(y, z) \ge 2 m_C$. Then $$ d(x_0, y) + d(x_0, z) \ge d(y, z) \ge 2 m_C $$ which implies that either $d(x_0, y) > m_C$ or $d(x_0, z) > m_C$. In the first case, we have $y \notin U_{m_C}$ and in the second, $z \notin U_{m_C}$. So the subcollection indexed by $C$ doesn't contain one of $y, z$ even though $y, z \in K$ so that $\{ U_n \}_{n \in C}$ is not a subcover of $K$. Hence, there is no finite subcollection of $\{ U_n \}_{n \in \mathbb{Z}_{> 0}}$ which covers all of $K$, so we conclude that $K$ is not compact. This proves the necessity of boudnedness for compact subsets of metric spaces. $\blacksquare$
Proposition: Let $X$ be a Hausdorff topological space. Then all compact subsets are closed.
Proof. Let $K \subseteq X$ be compact. Fix $x \notin K$, and for each $y \in K$ write $U_y, V_y$ to be disjoint open neighborhoods of $x, y$ respectively. Then we have an open cover $$ K \subseteq \bigcup_{y \in K} V_y $$ By compactness, there is a finite subcollection of $\{ V_y \}_{y \in K}$ indexed by $y_1, y_2, \cdots, y_n \in K$ for some $n \in \mathbb{Z}_{> 0}$ which still covers $K$. Then we define $U = \bigcap_{i = 1}^n U_{y_i}$. Clearly, this is an open neighborhood of $x$. Moreover, if $z \in K$ then there is $k \in \{ 1, 2, \cdots, n \}$ such that $z \in V_{y_k}$ so that $z \notin U_{y_i}$ and so $z \notin U$. That is to say, $U$ is disjoint from $K$ which means that $U \subseteq X - K$. So we have found an open neighborhood for an arbitrary $x \in X - K$ which is still contained in $X - K$. This proves that $X - K$ is open, and hence $K$ is closed which completes the proof.$\blacksquare$
These two facts together give one implication of the Heine-Borel theorem:
Corollary: The compact subsets of $\mathbb{R}^n$ are closed and bounded.
Proposition: Let $X$ be a topological space, $K \subseteq X$ compact and $L \subseteq K$ closed. Then $L$ is compact.
Proof. Let $\{ U_i \}_{i \in I}$ be an open cover of $L$ indexed by $I$. Then we have an open cover of $K$ $$ K \subseteq \bigcup \{ U_i, X - L \mid i \in I \} $$ By compactness of $K$, there is a finite subset $J \subseteq I$ such that $$ L \subseteq K \subseteq \bigcup \{ U_i, X - L \mid i \in J \} $$ Since $L, X - L$ are disjoint, we can safely remove the $X - L$ from the finite cover on the right and still cover all of $L$. So $\{ U_i \}_{i \in J}$ is a finite subcollection of the cover $\{ U_i \}_{i \in I}$ which covers $L$. Thus, we conclude that $L$ is compact. $\blacksquare$
Proposition: If $N \in \mathbb{Z}_{> 0}$ then $[-N, N] \subseteq \mathbb{R}$ is comapct.
Proof. Let $\{ U_i \}_{i \in I}$ be an open cover of $[-N, N]$. Let $$ C = \left\{ x \in \mathbb{R} \mid [-N, N] \cap (-\infty, x] \text{ admits a finite subcover from $\{ U_i \}_{i \in I}$} \right\} $$ Clearly, $-N \in C$ so $C$ is non-empty. Furthermore, if $x < y$ and $y \in C$ then also $x \in C$. This is because $[-N, N] \cap (-\infty, x] \subseteq [-N, N] \cap (-\infty, y]$ so the same finite subcover will suffice. Likewise, if $N \le x < y$ and $x \in C$ then also $y \in C$ because in this case $[-N, N] \cap (-\infty, x] = [-N, N] \cap (-\infty, y]$. These facts prove that either $C \subseteq (-\infty, N)$ or $C = \mathbb{R}$. We want to prove that the second case is true. Suppose for contradiction that the first case holds.
Then $C$ is bounded, and we can apply the least upper bound property of $\mathbb{R}$ to define $\eta = \sup \, C$. Then since $C \subseteq (-\infty, N)$, it follows that $\eta \le N$. Since $-N \in C$ we have also $\eta \ge -N$. Going back to the cover $\{ U_i \}_{i \in I}$, there exists $i_0 \in I$ such that $\eta \in U_{i_0}$ since $\eta \in [-N, N]$. Let $\varepsilon > 0$ be such that $(\eta - \varepsilon, \eta + \varepsilon) \subseteq U_{i_0}$. By properties of the supremum, there eixsts $y \in (\eta - \varepsilon, \eta]$ such that $y \in C$. By definition of $C$, we have that $[-N, N] \cap (-\infty, y]$ admits a finite subcover of $\{ U_i \}_{i \in I}$, so there is a finite $J \subseteq I$ such that $$ [-N, N] \cap (-\infty, y] \subseteq \bigcup_{i \in J} U_i $$ But then, $$ [-N, \eta] \subseteq \left( [-N, N] \cap (-\infty, y] \right) \cup (y, \eta] \subseteq \left( [-N, N] \cap (-\infty, y] \right) \cup (\eta - \varepsilon, \eta + \varepsilon) \subseteq \left( \bigcup_{i \in J} U_i \right) \cup U_{i_0} $$ which is a finite subcollection of $\{ U_i \}_{i \in I}$ which covers $[-N, N] \cap (-\infty, \eta]$. This proves that $\eta \in C$. But the same cover above also covers $[-N, N] \cap (-\infty, \eta + \varepsilon/2]$, so also $\eta + \varepsilon/2 \in C$. This contradicts $\eta$ being an upper bound of $C$.
We are forced to conclude that $C$ cannot be bounded, so $C = \mathbb{R}$. So $N \in C$ meaning that $[-N, N]$ admits a finite subcover from $\{ U_i \}_{i \in I}$. We can do this procedure for any open cover, so this proves that every open cover of $[-N, N]$ admits a finite subcover. Thus, $[-N, N]$ is compact. $\blacksquare$
Now recall the following theorem:
Theorem: Let $X, Y$ be topological spaces, and $A \subseteq X$, $B \subseteq Y$ compact subsets. Then $A \times B$ is compact in $X \times Y$ endowed with the product topology.
From this we have that for any $N \in \mathbb{Z}_{> 0}$, the box $[-N, N]^n \subseteq \mathbb{R}^n$ is compact. Finally,
Corollary: Closed and bounded subsets of $\mathbb{R}^n$ are compact.
Proof. Let $K \subseteq \mathbb{R}^n$ be closed and bounded. Boundedness implies that there exists $N \in \mathbb{Z}_{> 0}$ such that $K \subseteq [-N, N]^n$. Then since $K$ is a closed subset of a compact set, it must be compact itself. $\blacksquare$