211 views

### Why is the closed unit interval compact? [duplicate]

Looking at the definition for compactness, i.e. "every open cover has a finite subcover", it seems like $[0,1]$ wouldn't be compact, since you can't construct a cover with a finite subcover that fills ...
133 views

### Using the “open cover” definition of compactness to show that $[0,1]$ is compact [duplicate]

How can I prove that the interval $[0,1]$ is compact in real line topology? I know how to prove it using concept of boundedness and closedness, but I wish to understand it by using the "open cover" ...
16k views

2k views

### Compactness of the closed interval [0,1]

In general topology, a topological space is said to be compact, if every one of its open cover has a finite subcover. However, I cannot see the compactness of the close interval [0,1] from the above ...
480 views

### Compactness of the open and closed unit intervals

In the article by Tao it's explained that the compactness can be formulated in the most general way as: (All open covers have finite subcovers) If $V_\alpha:\alpha\in\mathcal{a}$ is any collection ...
### Finite cover for $[0,1]$
$F=[0,1] \subset \bigcup B_{r_j}(x_j)$ where $\{x_j\}$ is an arbitrary enumeration of rational numbers in $[0,1]$. $[0,1]$ is compact and thus must have a finite cover. $B_{r_j}(x_j)$ is an open ball ...