To show that the open interval $(0,1)$ is not compact using the covering argument, we need to demonstrate that it fails the Heine-Borel Theorem in $\mathbb{R}$, which states that a subset of $\mathbb{R}$ is compact if and only if it is both closed and bounded.

Step-by-Step Explanation:

1. Boundedness:

   • The interval $(0,1)$ is bounded because there exist real numbers $0$ and $1$ such that every point $x \in (0,1)$ satisfies $0 < x < 1$.

2. Openness vs. Closedness:

   • The interval $(0,1)$ is not closed because it does not contain its boundary points $0$ and $1$.
   • A compact set must be closed, so this already suggests that $(0,1)$ might not be compact. However, we must confirm this by considering an open cover.

3. Open Cover Definition:

   • An open cover of a set $S$ is a collection of open sets such that the union of these sets contains $S$.
   • A set is compact if every open cover has a finite subcover (i.e., a finite collection of those open sets that still covers $S$).

4. Construct an Open Cover:

   • Consider the following open cover of $(0,1)$: $\mathcal{U} = \left\{ \left(\frac{1}{n}, 1 - \frac{1}{n}\right) \, : \, n \in \mathbb{N}, n > 1 \right\}$
     • Each set $\left(\frac{1}{n}, 1 - \frac{1}{n}\right)$ is open, and as $n$ increases, these intervals get closer and closer to $(0,1)$.
     • The union of all these intervals is: $\bigcup_{n=2}^{\infty} \left(\frac{1}{n}, 1 - \frac{1}{n}\right) = (0,1)$
     • Thus, $\mathcal{U}$ is an open cover of $(0,1)$.

5. Check for a Finite Subcover:

   • Suppose we try to find a finite subcover of $\mathcal{U}$.
   • Any finite number of sets from $\mathcal{U}$ will leave a gap near $0$ or near $1$ because for any finite choice of $n$, there will always be points in $(0,1)$ that are not covered by the chosen intervals.
   • For example, if you take $\left(\frac{1}{n}, 1 - \frac{1}{n}\right)$ for a finite number of $n$, the interval will miss points close to $0$ and points close to $1$.

6. Conclusion:

   • Since there is no finite subcover that can cover the entire interval $(0,1)$, the interval fails the open cover criterion for compactness.
   • Therefore, $(0,1)$ is not compact.

Summary:

The open interval $(0,1)$ is not compact because we can construct an open cover that does not admit a finite subcover. This is a consequence of the fact that $(0,1)$ is not closed and does not contain its boundary points.

---

Do you want more details on this?

Here are some questions to consider:

1. What is the Heine-Borel Theorem, and how does it relate to compactness in $\mathbb{R}^n$?
2. How can we show that $[0,1]$ is compact using the same covering argument?
3. Why is boundedness important for compactness in $\mathbb{R}$?
4. Can you provide other examples of sets that are not compact in $\mathbb{R}$?
5. How would this proof change in higher dimensions, like $(0,1)^2$?

Tip: A useful intuition for compactness is that a compact set behaves like a "finite" set in terms of covering properties, despite possibly being infinite in size.