Compactness of [1,2] and [1,2).

Question

This is the definition I use for compactness: Let $X$ be a topological space. We say that $K \subset X$ is a compact set if:

1. $K$ is not empty
2. For any arbitrary open sets $U_i \subset X_{i∈I}$ whose union contains $K$, one can find a finite number of these open sets such that their union contains $K$

I want to demonstrate that $K_1 = [1,2]$ is compact and $K_2 = [1,2)$ is not, being $X=\mathbb{R}$ equipped with the usual topology.

For both cases I start saying that the space of all open covers $S_i$ exists for each $K_i$. Then I build a set $U_i = (0,2 - 1/i), i > 0$ and build a collection of sets $A_n = \{ U_i \mid 1 \le i \le n \}$

• For $K_2$, I recognize that:
  1. For any $x \in K_2$, I can always find a collection (of sets) $A_N$ such that the union of its members $U_i$ contains such $x$, so basically $K_2 \subset\bigcup U_i \in S_2$ .
  2. If I try to join a finite amount of $U_i$, say up to $i = M$, I can find a $x \in K_2$ such that $x \notin \bigcup_{i=1}^{M} U_i$.
  3. Then, I have found a collection of open sets whose union contains $K_2$ (1) but it is impossible to find a finite collection of those sets whose union still contains $K_2$ (2).
• For $K_1$, I recognize that I can find $x$ such that I would never find it on any $A_n$, this is $x=2$. If I build the collection of open sets $A'_i = \{U_i ,B_\epsilon(2) \mid i > 1\}$, where $B_\epsilon(1), \epsilon > 0$ is a (open) ball around $2$ of radius $\epsilon$.
  1. First I show that it is possible to cover $K_1$. For any $x \in K_1$, if $x = 2$, then $x\in B_\epsilon (2) \subset A'_i$ for any $A_i$. If $1 \le x < 2$ one can always find a collection (of sets) $A_N$ such that their union of its member $U_i$ contains such $x$, so, for any $x \in K_1$, it's possible to find a collection (of sets) $A'_N$ whose union of its members contains $K_1$: \begin{eqnarray} K_1 \subset \left(\bigcup U_i \right) \cup B_\epsilon \in S_1, \text{for any $\epsilon$} \end{eqnarray}
  2. Now we prove that it is not necessary to have all $U_i$. Given $M$, and $x \in K_1$, one can find a $x \notin \bigcup_{i=1} ^ M U_i$ and for any $x \notin \bigcup_{i=1}^M U_i$ there is a $B_\epsilon (2)$ such that $x \in B_\epsilon (2)$ for some $\epsilon \ge d$. Basically, I can find a finite collection of open set $A'_N =\{A_N, B_d(2)\}$ whose union of its memebers contains $x$
  3. We've shown that for any arbitrary collection of open sets (is this really arbitrary?) such that their union contains $K_1$, it is possible to find a finite collection of those open sets such that their union still contains $K$.

I want to know if there are mistakes, specially in the case for $K_1$.

Answer 1

Your proof that $K_2$ isn't compact is correct. But your proof of the fact that $K_1$ is compact isn't, because you did not start with an arbitrary family of open sets.

Answer 2

The idea for $K_2$ is right, but you should improve the presentation.

The family $\{U_n:n>0\}$, where $U_n=(0,2-1/n)$, is an open cover for $K_2$, because for every $x\in[1,2)$ there exists $n$ with $0<x<2-1/n$ (easy proof).

On the other hand, no finite subfamily is a cover, because if we are given $U_{n_1},U_{n_2},\dots,U_{n_k}$, we have $$ \bigcup_{i=1}^k U_{n_i}=U_n $$ where $n=\max\{n_1,\dots,n_k\}$ and $[1,2)$ is not a subset of $U_n$ for every $n$.

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Your proof for $K_1$ is unfortunately wrong. You need to start from an arbitrary open cover and are not allowed to choose your preferred one.

For instance, just take $\mathbb{R}$, which by itself is an open cover for $K_1$ and obviously has a finite subcover. Can you conclude from this that $K_1$ is compact? No, it would apply to every subset of $\mathbb{R}$.

The difference with the $K_2$ case is that for this set you want to show it isn't compact, so it's sufficient to find an open cover having no finite subcover.

The proof that $K_1$ (as well as any closed an bounded interval) is compact is quite a deep result,