# Math Help - Compact

1. ## Compact

Let K be a suet of R1 consisting of 0 and 1/n for n = 1,2,3,...Prove that K is compact directly from the definition without uing the Heine Borel Property.

I am not sure if my solution to this is right. I was given a hint to use the Archimedian property of Reaal numbers. Can you tell me if my solution is right?

Let {Gi} be an open cover for K. Then one of these open sets should cover 1, call it G1. Since it is an open set in R1, it is an open interval. Then we can find a h>0, (where h< radius of G1) such that (1-h,1) is a subset of G1.

By the Archimedian property, there exist only finitely many n such that
n(1-h)>1
ie) only finitely many n such that , (1-h) > 1/n

So these finitely many n can be covered by finite open sets with together with G1 and the open set covering 0 form the finite subcover of {Gi}.

Am I right?

2. Say that the collection $\left\{G_n\right\}$ is an open covering of $K$.
Then each $G_n=(\alpha_n,\beta_n)$, i.e an open interval.
$\left( {\exists a} \right)\left[ {0 \in G_a = \left( {\alpha _a ,\beta a } \right)} \right]$ so that $\alpha_a<0<\beta_a$.
$\left( {\exists N \in \mathbb{Z}^ + } \right)\left[ {n > N \Rightarrow \frac{1}{n} < \beta _a } \right]$.
$1 \leqslant k \leqslant N \Rightarrow \quad \left( {\exists n_k } \right)\left[{\frac{1}{k} \in G_{n_k } } \right]$
Now there is a finite subcover of $K$.

3. Originally Posted by poorna
Let K be a suet of R1 consisting of 0 and 1/n for n = 1,2,3,...Prove that K is compact directly from the definition without uing the Heine Borel Property.

I am not sure if my solution to this is right. I was given a hint to use the Archimedian property of Reaal numbers. Can you tell me if my solution is right?

Let {Gi} be an open cover for K. Then one of these open sets should cover 1, call it G1. Since it is an open set in R1, it is an open interval. Then we can find a h>0, (where h< radius of G1) such that (1-h,1) is a subset of G1.

By the Archimedian property, there exist only finitely many n such that
n(1-h)>1
ie) only finitely many n such that , (1-h) > 1/n

So these finitely many n can be covered by finite open sets with together with G1 and the open set covering 0 form the finite subcover of {Gi}.

Am I right?
It's true in any metric space (and any topological Hausdorff space for that matter) that if $\{x_n\}$ is a convergent sequence such that $x_n\to x$ then $C=\{x\}\cup\left\{x_n:n\in\mathbb{N}\right\}$ is compact. The same idea applies as Plato pointed out. The fact that any open set containig $x$ contains all but finitely many values of $\left\{x_n:n\in\mathbb{N}\right\}$