Relatively open sets problem

Question

I am trying to solve the following exercise:

Let $M, N, U \subseteq \mathbb C$ with $M \cup N = U$. The sets $M$ and $N$ are $U$-closed. Show that a set $V \subseteq U$ is open in $U$, if, and only if, $V\cap M$ is $M$-open and $V\cap N$ is $N$-open.

I thought about this for about 2 hours now but can't get to a solution. My thoughts:
For the "$\Rightarrow$"-direction I first assume that $x\in V\cap M$
I know that since $V$ is open in $U$ there is $\varepsilon > 0$ with $U_\varepsilon (x) \subseteq V$.
Furthermore, since $M$ is $U$-closed it holds that $U_\varepsilon(x)\cap (U \setminus M) \neq \emptyset$
Now let's assume $V\cap M$ is closed in $M$, $x\in \partial (V\cap M) $and try to get a condradiction.
Since $V\cap M$ is closed in $M$ it holds that $U_\varepsilon(x)\cap M\setminus(V \cap M) \neq \emptyset$. And here is where I don't know what to do. I feel like I am really close .

Answer 1

I have a feeling that the statement is overcomplicated just to blur the solution. Note that none of that depends on $\mathbb{C}$. So let's simplify it:

Let $M, N$ be closed subsets of $X$ such that $X=M\cup N$. Prove that $U$ is open in $X$ if and only if $U\cap M$ is open in $M$ and $U\cap N$ is open in $N$.

"$\Rightarrow$"

Obviously if $U$ is open in $X$ then it is open in subspaces due to the definition of subspace topology.

"$\Leftarrow$"

If $U\cap M$ is open in $M$ then $F=M\backslash(U\cap M)$ is closed in $M$. But $M$ is closed in $X$ and thus $F$ is closed in $X$ (closed set in a closed subspace is closed in whole space).

Analogously $F^{'}=N\backslash(U\cap N)$ is closed in $N$ and thus closed in $X$.

Therefore $F\cup F^{'}$ is closed in $X$. Now

$$F\cup F^{'}=(M\backslash(U\cap M))\cup (N\backslash(U\cap N))$$

The right side is equal to $X\backslash U$ because $M\cup N=X$. In particular $X\backslash U$ is closed and thus $U$ is open.

Answer 2

Here's metrical solution.

As before I'll simplify the statement a bit:

Let $M, N$ be closed subsets of a metric space $X$ such that $X=M\cup N$. Prove that $U$ is open in $X$ if and only if $U\cap M$ is open in $M$ and $U\cap N$ is open in $N$.

"$\Rightarrow$"

Let $x\in U\cap M$ (without loss of generality we can only consider $M$, proof for $N$ is exactly the same). We need to show that some ball in $M$ around $x$ is a subset of $U$. Since $U$ is an open subset of $X$ then there exists an open ball $B$ (in $X$) such that $B\subseteq U$. But $B\cap M$ is an open ball in $U\cap M$ because by definition $U\cap M$ has the same metric. This proves first implication.

"$\Leftarrow$"

We will show that $X\backslash U$ is closed in $X$. Let $(x_n)\subseteq X\backslash U$ be a convergent sequence. We will show that the limit $\lim x_n\in X\backslash U$.

Assume this is not true. Since $X=M\cup N$ then there exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $(x_{n_k})\subseteq M$ or $(x_{n_k})\subseteq N$. Let $(b_n)=(x_{n_k})$. Without loss of generality we may assume that $(b_n)\subseteq M$ and so $\lim b_n = \lim x_n \in M$ because $M$ is closed. But we've assumed that $\lim x_n\not\in X\backslash U$, i.e. $\lim x_n\in U$. Since $U\cap M$ is open in $M$ then there exists a ball $B$ such that:

1. $B$ is open in $M$
2. $B$ is around $\lim x_n=\lim b_n$
3. $B\subseteq U\cap M$

Since $b_n$ converges to its limit and $(b_n)\subseteq M$, then almost all elements of $(b_n)$ are in $B$. In particular $b_j\in B$ for some $j$. But $B\subseteq U\cap M$ so $b_j\in U$. Contradiction since we assumed that $(x_n)\subseteq X\backslash U$.