Are all covers fundamental?

Question

While attempting an exercise I proved something which I don't think is right, but i can't really find the error.

We say a family $R$ of subsets of a topological space $X$ is a cover if $\bigcup\{ A: A \in R\} = X$. We say a cover is fundamental if $U \subseteq X$ is open $\iff U \cap A$ is open in $A$ with the subspace topology $ \ \forall A \in R$.

The exercise stated the following: if $R$ is a cover such that $\bigcup\{ Int(A): A \in R\} = X$, then prove that $R$ is fundamental.

I thought i had something before realizing that i didn't really use hypothesis, so here's my proof that all covers are fundamental:

What we really want to prove is the $\Leftarrow$, since the $\Rightarrow$ follows from the definition of subspace topology. So, let $U \subseteq X: U \cap A_i$ is open in $A_i \forall A_i \in R$. By definition, $U\cap A_i$ is open $\iff \exists$ an open $ V_i \subseteq X: V_i \cap A_i = U \cap A_i$. Since $R$ is a cover, $X = \bigcup_i\{ A_i: A_i \in R\} \implies U = X \cap U = \bigcup_i\{ A_i: A_i \in R\} \cap U = \bigcup_i\{ A_i \cap U: A_i \in R\} = \bigcup_i\{ V_i \cap A_i: A_i \in R\} = (\bigcup_i\{V_i\}) \cap (\bigcup_i\{ A_i: A_i \in R\}) = (\bigcup_i\{V_i\}) \cap X = (\bigcup_i\{V_i\})$ which is open since it's the union of open sets, therefore $U$ is open and $R$ is a fundamental cover.

I never really used the fact that $\bigcup\{ Int(A): A \in R\} = X$, so where was i wrong? Also, could you give me some hints as to how to solve the exercise?

Answer 1

The mistake is in the set-theoretic operations you performed : Given two families $(C_i), (D_i)$, it is not necessarily the case that $\bigcup_i (C_i\cap D_i) = (\bigcup_i C_i)\cap (\bigcup_i D_i)$ : in the second term you have some undesired $C_i\cap D_j$ terms.

To be concrete, take $I=\{0,1\}$, $C_i\cap D_i = \emptyset$, $C_0= D_1, C_1 = D_0$.

Then the LHS is $\emptyset \cup \emptyset$ while the RHS is $(C_0\cup C_1)\cap (D_0\cup D_1) = C_0\cup C_1$ which we may of course choose to be nonempty.

Now to see an explicit example of the topological result fails, take $R$ to be the family of all singletons of $X$ : $R=\{\{x\}, x\in X\}$. This is quite clearly a cover of $X$. Now $U\cap \{x\} = \emptyset$ or $\{x\}$ for all $x\in X$ so $U\cap \{x\}$ is always open in $\{x\}$ : if $R$ were fundamental, it would mean that $X$ is discrete. Thankfully, there exist nondiscrete spaces.

In this example, for $x\in R$, you may choose $V_x = X$ or $\emptyset$ depending on whether $x\in U$ so that the LHS is precisely $U$ and the RHS is $X$ : you had unwanted $C_i\cap D_j$'s !

Answer 2

I think i figured the exercise out by the way, i'm posting for anyone who is interested in it. That was not that difficult in the end.

Lemma: let $X$ be a topological space, $A \subseteq X$. If $E \subseteq A$ is open in $A$ with the subspace topology, then $E \cap Int(A)$ is open in $X$.

Proof of lemma: $E \subseteq A$ is open $\iff \exists$ an open $ V \subseteq X: E=V \cap A$. We have $E \cap Int(A) = E \cap A \cap Int(A) = V \cap Int(A)$. Since both $V$ and $Int(a)$ are open, their intersection is open. ∎

Solution of exercise: we want to prove that if $R$ is a cover such that $\bigcup\{ Int(A): A \in R \}=X$ and $U \subseteq X$ is such that $U \cap A$ is open in $A \ \forall A \in R$, then $U$ is open in $X$. Since $U=U \cap X$, we have $U=U \cap (\bigcup\{ Int(A): A \in R \}) = \bigcup\{ Int(A) \cap U: A \in R \}$. By the lemma, we know that if $U \cap A$ is open in $A$ then $U \cap Int(A)$ is open in $X$, so we have written $U$ as the union of open sets which means it is, indeed, open.∎