# Proving '$A$ open in $V\subseteq M$ (metric space) iff $A=C\cap V$ (certain open $C$ in $M$)'

I want to prove the following:

Let $(M,d)$ be a metric space. Let $A\subseteq V\subseteq M$.

1) $A$ is open in $V \Leftrightarrow A = C\cap V$ (for a certain open $C$ in $M$)

2) $A$ is closed in $V \Leftrightarrow A = C\cap V$ (for a certain closed $C$ in $M$)

### Questions:

• Could someone check the proof?

• 'for a certain open $C$ in $\color{Blue}{M}$.'

Would this proof also work for a more specific choice of $C$? Like for a certain open $C$ in $\color{blue}{V}$. I don't really see the added value of choosing $M$ over $V$.

• Could some give me some pointers on how to prove $2, \Rightarrow$?

### Proof 1)

$\Leftarrow$: Choose $a\in A$.

$$\begin{array}{rl} & a \in A = C\cap V\\ \Rightarrow & a \in C\\ \Rightarrow & (\exists r > 0)(B_M(a,r)\subseteq C)\\ \Rightarrow & (\exists r > 0)(B_M(a,r)\cap V \subseteq C\cap V)\\ \Rightarrow & (\exists r> 0) (B_V(a,r)\subseteq A \end{array}$$

$\Rightarrow$: Choose $a\in A$.

$$\begin{array}{rl} \Rightarrow & (\exists r_a >0)(B_V(a,r_a) \subseteq A) \end{array}$$

Consider all $a\in A$ then:

$$\begin{array}{rl} & A = \bigcup_{a\in A} B_V(a,r_a)\\ \Rightarrow & A = \bigcup_{a\in A} \left[ V\cap B_M(a,r_a)\right]\\ \Rightarrow & A = V\cap\left[ \bigcup_{a\in A} B_M(a,r_a)\right] \end{array}$$

Let $$\left[ \bigcup_{a\in A} B_M(a,r_a)\right] = C$$ which is open as a union of open sets.

### Proof 2)

$\Leftarrow$:

$$\begin{array}{rrl} & V\setminus A &= V\setminus(C\cap V)\\ \Rightarrow & & = (V\setminus C)\cup (V\setminus V)\\ \Rightarrow && = V\setminus C \end{array}$$

Since $C$ is closed then $V\setminus C$ is open and so is $V\setminus A$. Then $A$ is closed in $V$.

$\Rightarrow$: How?

So far everything seems right. For the last proof, let $A$ closed in $V$. Then $V\backslash A$ is open in $V$. By the first statement (which you already proved) there exists an open $B$ in $M$ such that $V\backslash A=B\cap V$. Now $A=(M\backslash B)\cap V$. And we know that $C= M\backslash B$ is closed in $M$ because $B$ is open in $M$.

Sidenote: For non-metric spaces, we actually define the induced topology (the most natural and conventional one) on a subset of a topological space like this.