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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!

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  • $\begingroup$ "Let $k \in X\cup Y$. [...] Therefore $\exists k \in X$" You're using $k$ for two seemingly different points here. $\endgroup$ – Arthur May 23 at 7:16
  • $\begingroup$ @Arthur let me correct that! Thanks $\endgroup$ – Bruno Reis May 23 at 7:17
  • $\begingroup$ 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)! $\endgroup$ – user May 23 at 7:56
  • $\begingroup$ @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! $\endgroup$ – Bruno Reis May 23 at 8:03
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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 $k=a$?

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  • $\begingroup$ Now you got me... for $k \notin \{a,b\}$ that is true... But can I guarantee that there is such $k \in X$? $\endgroup$ – Bruno Reis May 23 at 7:22
  • $\begingroup$ @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. $\endgroup$ – 5xum May 23 at 7:24
  • $\begingroup$ Hmmm... So I just need to treat those cases and we're done right? $\endgroup$ – Bruno Reis May 23 at 7:27
  • $\begingroup$ @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. $\endgroup$ – 5xum May 23 at 7:29

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