Let $X$ a set not empty and $(X,d)$ a metric space. Prove he union of two open sets is open.
My proof:
Let $A_1,A_2$ open sets, we need to prove $A_1\cup A_2$ is open.
As $A_1,A_2$ are open set, then for all $a_1,a_2\in A_1,A_2$ respectively we have $r_1,r_2>0$ such that $B(a_1,r_1)\subset A_1$ and $B(a_2,r_2)\subset A_2$
Let $r=\frac{1}{2}min\{r_1,r_2\}$ and $x\in A_1\cup A_2$, then $x \in A_1$ or $x \in A_2$
This implies: $B(x,r)\subset A_1\cup A_2$
In conclusion, $A_1\cup A_2$ is open set.
Note: My definition of Open is $\forall x\in A_1$ exists $r>0$ such that $B(x,r)\subset A$
What is your opinion about my proof? Do you think is a good proof? Is convincing?
Best Answer
Nope. It isn't a good proof. Note that $r_1, r_2$ depends on the elements $a_1,a_2$. There is a simpler proof: Take $x \in A_1 \cup A_2$. Then $x \in A_1$ or $x \in A_2$. If $x \in A_1$, as $A_1$ is open, there exists an $r>0$ such that $B(x,r) \subset A_1 \subset A_1 \cup A_2$ and $B(x,r)$ is an open set.
Therefore $A_1 \cup A_2$ is an open set.