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Is $B((-2,0);1) \cup B((2,0);1 \cup \{(x,0): -1<x<1\}$ compact in the space $\mathbb{R}^2$ with Euclidean metric.

$B((-2,0);1)$ reads as an open ball centered at $(-2,0)$ with radius 1.

I was told that I must show that each region is not compact. My question is why must I do each region individually and not as an entire set?

My first attempt is as follows:

Let $S = B((-2,0);1) \cup B((2,0);1 \cup \{(x,0): -1<x<1\}$, and let $$A_n = B\left((0,0);3-\frac{1}{n}\right).$$ Then clearly $$S \subset\bigcup_{n\in\mathbb{N}}A_n,$$ so it is an open cover of S. I know how to show it does not contain a finite subcover that covers S. But why can't I show that the space is not compact in one go?

SOULed_Outt
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1 Answers1

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The union is not closed. Consider the point $(1,0)$.

Thus it is not compact.

"My question is why must I do each region individually and not as an entire set?"

The answer is: no, you don't need to do so.

  • Right. But I think we are to prove that is not compact by finding covers. I was told to find an open cover that has no finite subcover for each set in the union. Why can't I just find an open cover with no finite subcover for the entire union? – SOULed_Outt Oct 10 '16 at 01:20
  • Sure you can. But how do you find it directly? –  Oct 10 '16 at 01:31
  • With a ball centered at the origin with a radius of $3-\frac{1}{n}$ for $n\in\mathbb{N}$ – SOULed_Outt Oct 10 '16 at 01:32
  • Well, this would work. –  Oct 10 '16 at 01:37
  • Okay. I was told to consider each set in the union individually. I gave some though behind that recommendation. Would a compact set union a non-compact set be a compact set altogether? – SOULed_Outt Oct 10 '16 at 01:40