To prove that if $K \subset X$ is compact in a metric space $X$ that is a union of open neighborhoods, then $K$ is closed and bounded, we will tackle each property (closed and bounded) separately.

1. Boundedness

We need to show that the compact set $K$ is bounded. A set $K \subset X$ is bounded if there exists some real number $M > 0$ and a point $x_0 \in X$ such that:

$d(x_0, k) \leq M \quad \text{for all} \, k \in K.$

Proof of boundedness:

• Since $K$ is compact, by definition, every open cover of $K$ has a finite subcover.
• Consider the family of open balls $\{ B(x, n) \}_{n \in \mathbb{N}}$, where $B(x, n)$ denotes the open ball of radius $n$ centered at some point $x \in X$.
• These open balls cover the entire space $X$, which means that $\{ B(x, n) \}$ forms an open cover for $K$.
• By the compactness of $K$, there is a finite subcover, say $\{ B(x, n_1), B(x, n_2), \dots, B(x, n_m) \}$.
• Let $n_{\max} = \max\{n_1, n_2, \dots, n_m\}$. This implies that $K \subset B(x, n_{\max})$.
• Therefore, $K$ is contained within a ball of finite radius, meaning $K$ is bounded.

2. Closedness

We need to show that $K$ is closed, meaning that if a sequence $\{x_n\} \subset K$ converges to some limit $x \in X$, then $x \in K$.

Proof of closedness:

• Let $\{x_n\}$ be a sequence in $K$ that converges to some point $x \in X$.
• Since $K$ is compact, every sequence in $K$ has a convergent subsequence whose limit lies in $K$.
• But $\{x_n\}$ already converges to $x \in X$. By the compactness of $K$, this limit must also lie in $K$, implying that $x \in K$.
• Therefore, $K$ contains all its limit points, which means $K$ is closed.

Conclusion:

We have shown that if $K \subset X$ is compact, then:

• $K$ is bounded because it is contained in some large ball.
• $K$ is closed because every convergent sequence in $K$ has its limit point within $K$.

Thus, in a metric space $X$ that is a union of open neighborhoods, every compact set $K$ is closed and bounded.

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Would you like further details or have any questions? Here are some related questions to expand on this topic:

1. How does the Heine-Borel theorem relate to this result?
2. What are the definitions of open and closed sets in metric spaces?
3. Can you prove that every closed and bounded subset of $\mathbb{R}^n$ is compact?
4. What happens if the metric space is not a union of open neighborhoods?
5. How do compact sets behave in non-metric topological spaces?

Tip: In metric spaces, compact