Show that the set is compact using the definition

Question

The set in question is $\{0\}\cup \{1,\frac12,\frac13,\ldots,\frac1n,\ldots\}$ (for $n\in\mathbb N$).

Okay, so for a set to be compact, every open cover of it must be able to be broken down into a finite subcover.

An open subcover of $(-2, 2)$ for instance can be broken down into $(-2,0) \cup (0,2)$. Is my understanding of the definition correct?

Intuitively, I feel like every open set can't possibly be compact. Is this correct. What's the correct approach to putting the logic into mathematical language?

Answer 1

An open cover of $(-2,2)$ can only be broken down into $(-2,0) \cup (0,2)$ if those two open sets are included in the cover, not to mention that $(-2,0) \cup (0,2)$ fails to cover $(-2,2)$ to begin with (what contains $0$?).

You are right when you say that open sets of $\mathbb R$ are not compact (well the empty set is). To show this, you would have to construct a open cover which can never be reduced to a finite subcover. The simplest open set to look at for this would be $(0,1)$.

To address your main question, is a singleton, say $\{0\}$, compact? What about a set of two real numbers? Three? Some natural number $n$?

Answer 2

Let $A = \{0\} \cup \{\frac 1 n \ | \ n \in \Bbb N\}$.

An open cover of $A$ is a collection of open sets $\{G_{\lambda} \ | \ \lambda \in I\}$ such that the union of all $G's$, $$\bigcup_{\lambda \in I} G_{\lambda} \supseteq A$$ or in other words the union of $G_{\lambda}$'s covers $A$. Now the reason we use an index $\lambda$ and an index set $I$ is to make the collection $\{G_{\lambda}\}$ "as infinite as you like". That is the number of $G$'s in $\{G_{\lambda}\}$ can be an incredibly large, unfathomable number.

But now, we say the set $A$ is compact if given such an open covering of $A$, $\{G_{\lambda}\}$ contains a finite number of elements $\{G_i \ | \ i = 1,2,..., n\}$ such that the union formed by this finite sub-collection also covers $A$. That is there exists a subcollection $\{G_i \ | \ i = 1,2,..., n\} \subseteq \{G_{\lambda}\}$ such that $$\bigcup_{i = 1}^n G_i \supseteq A$$

is also true.

Now to prove $A$ is compact, you are required to assume that $\{G_{\lambda}\}$ is an arbitrary open cover of $A$ and then conclude there is a finite number of elements $\{G_1, G_2, ...G_n\}$ in $\{G_{\lambda}\}$ such that $\bigcup_{i = 1}^n G_i$ also covers $A$.

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Here are the things you need to know to prove that the set $A = \{0\} \cup \{\frac 1 n \ | \ n \in \Bbb N\}$ is compact.

• $\bigcup_{\lambda \in I} G_{\lambda} \supseteq A$ means for any $a \in A$ there exists $\lambda' \in I$ such that $a \in G_{\lambda'}$. And this is true for every element in $A$, in particular $0$.
• A set $G_\lambda$is open if given any $x \in G_\lambda$ there exists an open ball around $x$ which is also contained in $G_\lambda$. That is given any $x \in G_\lambda$ there exists $r \gt 0$ such that $\{y \ | \ |y - x| \lt r\} \subseteq G_\lambda$

Give it a shot. Drop in a comment if you need further help.

Answer 3

Following Ishfaaq, I just want to check if I get his hint since it has been $8$ years now.

Since ${0}$ is the limit of this sequence, it is very essential and powerful to solve the proof. The idea is that although the sequence is infinite approaching $0$, there must ONE set that contains all the infinite elements $\{\frac{1}{n}\}_{n>M}^{\infty}$. Why? Note the open set definition, i.e., the neighborhood of $0$ should be also contained in the set that contains $0$.

We must have $\exists G_i, 0\in G_i$. For this $G_i$, assume that there exists a positive value $b\in G_i$, otherwise, it will not be an open set.

• if $b\ge 1$, then $G_i$ will cover A.
• if $0<b<1$, then let $M=int(\frac{1}{b}+1)$, all elements $\{\frac{1}{n}\}_{n>M}^{\infty} \subseteq G_i$. Then all not-covered elements in A are finite thus we can have finite sets to cover them. Then plus $G_i$, still finite, and cover all elements.