Proving the open disc is not compact

Question

Prove that the open disc $D=\{(x,y):x^2+y^2<1\}$ considered as a subspace of $\mathbb{R}^2$.

I can find the following covering of the disc $B_{1-\frac{1}{n}}(0,0):n\in\mathbb{N}$.

$D\subseteq\bigcup_\limits{i=1}^{\infty}B_{1-\frac{1}{n}}(0,0)$.

This covering is infinite and there is no finite sub covering then $D$ is not compact.

Question:

Is this right? If not. What should I do to prove $D$ is not compact?

Thanks in advance!

Answer 1

You are right. But you have to show, that each collection of finitely many of your discs do not cover the set $D $.

Answer 2

You should show that $$D\not\subset B_{1-\frac{1}{m}}(0,0)$$

Therfore you should show that there is some $(x,y)\in D$ where $(x,y)\notin B_{1-\frac{1}{m}}(0,0)$.

For that you might just choose $(x,y)=(1-\frac{1}{m},0)$.

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alternatively

Use the theorem of Heine and Borel (sometimes referred to as Borel and Lebesgue) which says that in Euclidean space $\mathbb{R}^n$ the compact subsets are identical to the ones that are closed and bounded.

Answer 3

In $\mathbb{R}^2$, a compact set is closed and bounded...

Here, the set $D$ is not closed.