Linked Questions

0
votes
0answers
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 ...
0
votes
0answers
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" ...
121
votes
15answers
16k views

What should be the intuition when working with compactness?

I have a question that may be regarded by many as duplicate since there's a similar one at MathOverflow. The point is that I think I'm not really getting the idea on compactness. I mean, in $\mathbb{R}...
9
votes
4answers
19k views

Showing that $[0,1]$ is compact

Let's choose an open covering for $\left [ 0,1 \right ]$. For example $$\left \{ \left ( \frac 1 n,1-\frac 1 n \right ) \mid n\in \{ 3,4,\dots\} \right \}.$$ How can one choose a finite open ...
5
votes
4answers
4k views

Every open cover of $[0,1]$ has finite subcover

I am understanding proof of theorem stated in title from Spivak's calculus. It is as below. (0) Let $\mathcal{O}$ be an open cover of $[0,1]$. (1) Let $A=\{x\in [0,1]:[0,x] \mbox{ has finite ...
4
votes
5answers
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 ...
6
votes
3answers
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 ...
3
votes
2answers
427 views

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 ...
1
vote
2answers
2k views

Prove that $(a,b)$ is not compact based on the definition of compactness

I am trying to show that $(a,b)$ is not compact while $[a,b]$ is compact, purely based on definition of compactness. Here are examples, which I write it better: $[a,b]$ is covered by the intervals $(a-...
2
votes
4answers
183 views

Proof as to Why $[a, b]$ is Compact: Question About a Specific Step

I understand that the definition of compactness varies. I am wanting to prove the set $[a,b]$ is compact the following way just like Julien did here for an answer. This proof is also similarly set up ...
1
vote
3answers
731 views

Showing that the function $C(b)$ is a compact set for $|b| > 1$

I am reading "An Invitation to Dynamical Systems", and one of the challenge problems is to prove that $C(b)$ is a compact set where $C(b)$ is defined as the set of all numbers that can be expressed in ...
0
votes
2answers
342 views

Where to go on this proof involving closed intervals and unions of open intervals

I am absolutely stuck on this analysis problem from Taylor's Foundations of Analysis: "Prove that if $I$ is a closed, bounded interval which is contained in the union of some collection of open ...
-1
votes
3answers
81 views

Show that the set $[0,1]×[0,1]$ is compact in $\mathbb{R}^2$ to standard metric

I have to show that the set $[0,1]×[0,1]$ is compact in $\mathbb{R}^2$ with respect to the standard metric. I have to show this using only the definition of compactness. The definition I am given is:...
1
vote
1answer
77 views

A “Simple Chain of Regions” and Compactness in the Continuum

Let me just start by saying that I'm basically trying to prove this: How to prove every closed interval in R is compact? Except that I need to do it in a very strange way... I'm teaching an Inquiry-...
0
votes
2answers
55 views

Prove that $[7, 9]$ is covered by this family of intervals

For each $x \in\mathbb{R}$ let $r_{x} > 0$. Suppose $r_x$ is unspecified and can be different with each $x$. Consider the uncountable family of open intervals: $U = \{I_{r_x}(x): x \in\mathbb{R}\}$...

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