# $X,Y$ disjoint closed sets such that $X\cup Y = [a,b]$. Show that $X= \emptyset$ or $Y = \emptyset$

Let $$X,Y$$ be two disjoint closed sets on $$\mathbb{R}$$ such that $$X\cup Y = [a,b]$$. Show that $$X= \emptyset$$ or $$Y = \emptyset$$.

Here's what I've got by now:

Let $$k \in X\cup Y$$. Therefore $$k \in X$$ or $$k \in Y$$. Suppose that $$X \neq \emptyset$$ and $$k \in X$$, hence $$k \notin Y$$ which implies that $$k \in \mathbb{R} \setminus Y$$. From that it follows that $$\exists \epsilon_k > 0$$ such that $$(k - \epsilon_k, k + \epsilon_k) \subset \mathbb{R} \setminus Y$$. Since $$k \in X\cup Y = [a,b]$$ it follows that $$a \leq k \leq b$$.

Now let's get an $$\epsilon > 1$$ such that $$(k-\frac{\epsilon_k}{\epsilon},k+\frac{\epsilon_k}{\epsilon}) \subset \big((k - \epsilon_k, k + \epsilon_k) \cap [a,b] \big)$$. That $$\epsilon$$ clearly exists, and then it follows that $$(k-\frac{\epsilon_k}{\epsilon},k+\frac{\epsilon_k}{\epsilon}) \subset \mathbb{R}\setminus Y \cap [a,b]$$.

Now see that $$\mathbb{R} \setminus Y \cap [a,b] = \mathbb{R} \setminus Y \cap (X \cup Y) = X$$. From our previous conclusion, it follows that $$X$$ is open.

Using the same reasoning, by supposing that $$Y\neq 0$$ it follows that $$Y$$ is open.

So now suppose that $$X,Y \neq \emptyset$$. Theferore $$X,Y$$ are open sets. Therefore $$X\cup Y$$ is also an open set, hence it cannot be a closed interval $$[a.b]$$ which is closed, therefore $$X = \emptyset$$ or $$Y = \emptyset$$.

Edit for cases when $$k=a$$ or $$k=b$$

As the user 5xum pointed out, to guarantee the existence of that $$\epsilon > 0$$ we need $$k \notin \{a,b\}$$. So if I can guarantee that there is such $$k \in X$$, we're done without loss of generality for the case where $$Y \neq \emptyset$$.

To prove that, suppose that there isn't such $$k \in X$$. Therefore $$X=\{a\}$$ or $$X=\{b\}$$ or $$X=\{a,b\}$$. In all those three cases $$Y$$ cannot be closed, because $$Y$$ would be respectively $$(a,b]$$, $$[a,b)$$ and $$(a,b)$$. Since $$Y$$ is closed, that $$k$$ exists.

Can someone please check my work? That was a hard lemma for me and even after that proof attempt, I'm not 100% sure if it's fully correct! Thanks and any kind of help is highly appreciated!

• "Let $k \in X\cup Y$. [...] Therefore $\exists k \in X$" You're using $k$ for two seemingly different points here. – Arthur May 23 at 7:16
• @Arthur let me correct that! Thanks – Bruno Reis May 23 at 7:17
• Your proof is basically correct - you have all the right ideas. However, strictly speaking your conclusion isn't justified: you have shown that $X$ and $Y$ are open, hence $X \cup Y$ is open. But there exist sets which are both open and closed (e.g. $\mathbb R$). So you do need a small argument why a closed interval is not open. Perhaps this was obvious to you, but I thought I'd mention it because it's a common mistake to think sets that aren't open are closed and vice versa (there are also sets which are neither)! – user May 23 at 7:56
• @user Thanks a lot for pointing that out mate, but it was a previously proved fact that a closed interval is not open! I've also made an edit for the cases where $k \in \{a,b\}$ which would not guarantee the existence of $\epsilon$. Now I think I'm done! Thanks anyway! – Bruno Reis May 23 at 8:03

Now let's get an $$\epsilon > 1$$ such that $$(k-\frac{\epsilon_k}{\epsilon},k+\frac{\epsilon_k}{\epsilon}) \subset \big((k - \epsilon_k, k + \epsilon_k) \cap [a,b] \big)$$. That $$\epsilon$$ clearly exists
Does it though? What about if $$k=a$$?
• Now you got me... for $k \notin \{a,b\}$ that is true... But can I guarantee that there is such $k \in X$? – Bruno Reis May 23 at 7:22
• @BrunoReis Well, if such $k\in X$ doesn't exist, then $X$ is equal to either $\{a\}$, $\{b\}$ or $\{a,b\}$, and in all those three cases, $Y$ cannot be closed. – 5xum May 23 at 7:24
• @BrunoReis Yeah, apart from that, your proof looks OK to me. You should just change the "Let $x\in X\cup Y$" to "Let $x\in X$". This is because what you are proving at that point is that $X$ is open. – 5xum May 23 at 7:29