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!
Best Answer
Does it though? What about if $k=a$?