# Proving the union of two open sets is open using the general definition of metric space.

Prop: The union of two open sets is open.

Pf: Let $$(X,d)$$ be a metric space, and $$U$$ and $$V$$ are non empty and open sets such that $$U\subset X$$ and $$V \subset X$$. Let $$p$$ be a point such that $$d(p,q)< \epsilon_1$$ for some $$\epsilon_1 > 0$$ and $$q \in U$$. And $$d(p,q) for some $$\epsilon_2 >0$$ and $$q \in V$$. Since we know that $$q\in V$$ and $$q \in U$$, we notice $$q \in( U \bigcup V)$$. Therefore, $$q$$ is an interior point of the union of those sets $$\square$$.

I want to know if my proof is well-phrased and proves the proposition.

• Please explicitly state in your post what exactly it is that you are asking for. Presumably, you would like us to check your work. Is that correct? Is there anything else that you're looking for here? – Ben Grossmann Dec 13 '19 at 11:09

With all due respect, your proof is chaotic. For $$x \in X$$ and $$r>0$$ let us denote by $$B_r(x)$$ the set

$$B_r(x)= \{y \in X: d(y,x)

If $$U$$ and $$V$$ are open subsets of $$X$$, let $$W:=U \cup V.$$ We have to show that $$W$$ is open.

To this end take $$x \in W$$. Then $$x \in U$$ or $$x \in V.$$ WLOG we can assume that $$x \in U.$$ Then there is $$r>0$$ such that

$$B_r(x) \subseteq U.$$

Then we derive that

$$B_r(x) \subseteq W.$$

The proof is complete.

Note: since the asker has been updating their proof since originally posting the question, this answer may no longer address the proof in its current form.

Your proof could be substantially improved. Some criticisms below.

Let $$(X,d)$$ be a metric space, and $$U$$ and $$V$$ are non empty sets such that $$U\subset X$$ and $$V \subset X$$.

You should state that $$U$$ and $$V$$ are open sets. Also note that at some point, your proof should also address the case where either $$U$$ or $$V$$ is empty.

Then, $$U$$ and $$V$$ are also a metric space.

I have no idea what this sentence is supposed to mean; it should probably be removed.

We shall denote a point $$p$$ such that $$d(p,q)< \epsilon_1$$ for some $$\epsilon_1 > 0$$, then $$q \in U$$. And we can find a point $$p$$ such that $$d(p,q) for some $$\epsilon_2 >0$$, then $$q \in V$$. Since $$q\in V, q \in U \implies q \in( U \bigcup V)$$, $$q$$ is an interior point of the union of those sets $$\square$$

I don't understand this (which appears to be the core of your proof). At the very least, your first sentence "We shall denote a point $$p$$ such that $$d(p,q)< \epsilon_1$$ for some $$\epsilon_1 > 0$$, then $$q \in U$$" makes no sense as written.

Because of the definition of an open set in a metric space, your proof should be structured as follows: first, define what $$p$$ is. Say something like let $$p$$ be a point ("any point" or "an arbitrary point" also work) in $$U \cup V$$. Then, using the fact that $$U$$ and $$V$$ are open, show that there exists an $$\epsilon > 0$$ such that $$q \in U \cup V$$ whenever $$d(p,q) < \epsilon$$. This can be divided into parts:

• Define $$\epsilon$$
• State what $$q$$ is. Again, say something like "let $$q$$ be any point in $$X$$ with $$d(p,q) < \epsilon$$".
• Show that we have $$q \in U \cup V$$.
• @Melz you have done a good job of addressing the first two points and saying what $p$ is, but the rest of the proof still needs work. – Ben Grossmann Dec 13 '19 at 11:41