# Finding open covers that do not contain finite subcovers

I'm being asked that for each of the following spaces $(X_i, T_i)$, find an open cover $U_i$ that does not contain a finite subcover.

$X_i$ is a set and $T_i$ is a collections of subsets.

I have come up with a few answers that seem too easy and are now causing me to doubt my interpretation of the question.

I interpreted it as: find an open conver $U_i$ that contains an infinite amount of subcovers.

Here are the spaces and my answers:

1) $X_a = \mathbb{R}$, $T_a = \left \{(-\infty,a) : a \in \mathbb{R} \cup \left \{ \pm \infty \right \} \right \}$

My answer is the open set $(-n,n)$

• it is open
• it covers the collections of sets given
• it contains infinitely many subcovers (i.e- (-1,1),(-2,2),...))

2) $X_b = \mathbb{Q} \cap [0,1]$, $T_b = \left \{ U \cap [0,1] : U \subset \mathbb{R} \right \}$

Once again, my answer is $(-n,n)$ which seems to satisfy all conditions once again. For an open cover with finite subcovers I would have just picked $(-1,2)$

3) $X_c = \left \{ x \in \mathbb{R} : 0 < \left \| x \right \| \leq 1 \right \}, T_c$ is the standard topology

Here, the set is a circle with radius 1, surrounding the centre point in 2 dimensions.

Assuming the standard topology is simply all open covers $U \subset \mathbb{R^2}$, my answer becomes $(n_1,-n_1),(n_2,-n_2)$ which seems to satisfy all conditions again.

Can someone correct me if I'm wrong and help lead me to a proper understanding of the question.

Thanks, Greg.

1) $\left \{ (- \infty , a ) : a \in \mathbb{R} \right \}$

2) $\left \{ 0,1 \right \} \cup \left \{ (\frac{1}{n},1-\frac{1}{n}) : n \in \mathbb{N} \setminus \left \{ 1 \right \} \right \}$

3) $\left \{ \left \{ x \in \mathbb{R^2}: 0 < ||x|| \leq (1 - \frac{1}{n}) \right \} : n \in \mathbb{N} \setminus \left \{ 1 \right \} \right \}$

• For question 1), the set $(-n,n)$ are not open for the topology $T_a$, so $\{ (-n,n) | n \in \Bbb N \}$ is not an open cover of $X_a$.

• For question 2), the set $(-n,n)$ are not subset of $X_b$, so it cannot be an open cover. Even if you take the intersection with $X_b$, your cover $\{ (-n,n) \cap X_b | n \in \Bbb N \}$ has a finite subcover : $\{ (-2,2) \cap X_b \}$

• For question 3), same problem than with question 2).

Remember, given $(X,T)$ a topological space, an open cover of $X$ is a collection of subsets $U_i \in T$ such that $\bigcup_i U_i = X$

And you want to find open covers that, if you take only a finite number of elements in this cover, they don't cover $X$.

• this is random and irrelevant, but where you use your "such that" symbol in your sets, you can use \mid and it will look nicer. – Alex Mathers Nov 3 '15 at 1:17
• Ok, I think that makes sense now. See my Edit for updated answers. I'd like to know if I grasped it properly now. – Gregory Peck Nov 3 '15 at 13:26

Not a direct answer, but too long for a comment.

I'm being asked that for each of the following spaces $(X,T)$, find an open cover $\{U_{i}\}$ that does not contain a finite subcover.

Probably you're investigating this definition because of its relationship with compactness. Many students find the following adversarial game to help convey the meaning of "finding an open cover having no finite subcover".

• A referee supplies a topological space $(X, T)$.

• The first player selects a collection $\{U_{i}\} \subset T$ of open sets whose union is $X$. (The collection is said to be an open cover of $(X, T)$.)

• The second player attempts to select finitely many of the $\{U_{i}\}$ whose union is $X$. (If this can be done, the finite family forms a finite subcover (from the covering $\{U_{i}\}$).

If the second player succeeds (i.e., "wins the game"), the open cover $\{U_{i}\}$ has a finite subcover. Otherwise "the first player wins", and, the open cover $\{U_{i}\}$ has no finite subcover.

Your task is to find open covers for which the first player wins.

This "game" is mathematically interesting because it turns out there exists a large collection of topological spaces, called compact spaces, for which the second player wins against a perfect opponent; that is, for which every open cover of $(X, T)$ has a finite subcover.

The game is pedagogically interesting because humans are generally better at strategizing adversarial negotiations than they are at assimilating multiply-quantified sentences.

For example, it should be clear that if "the first player blunders" (for instance, taking $X$ itself as one of the $\{U_{i}\}$), then the second player wins. This, however, is not a property of the space $(X, T)$.

It's remarkable that infinite compact spaces exist at all: No matter how devious or perverse the first player is in picking open sets whose union is $X$, the second player can discard all but finitely many and still cover $X$.