I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect if there are parts that are overly wordy/complicated, or if I'm missing some element of proof that is helpful to see, if not strictly necessary.
The Prompt (from here):
Show that closed intervals in $\mathbb R^1$- sets of the form $\{x:a\leq x\leq b\}$ for fixed numbers $a$ and $b$ - are closed sets.
My Proof:
Suppose $I = \{x:a\leq x\leq b\}$ and is thus a closed interval in $\mathbb R^1$. Then $I^c =\{y\in\mathbb R^1\mid y<a \text{ or }y>b\}$ which we will split into two sets $I_a^c= \{y\in\mathbb R^1\mid y<a\}$ and $I_b^c= \{y\in\mathbb R^1\mid y>b\}$ so that $I_a^c \cup I_b^c=I^c.$ Let $y$ be an arbitrary element of $I_a^c$ and define $e=a-y.$ Now define open ball $B_e(y) = \{z\in\mathbb R^1\mid |z-y| < e\}$. For $z \in B_e(y)$ we can see: \begin{align*} |z-y| &< e\\ |z-y| &< a-y\\ z-y &< a-y\\ z &< a\\ \end{align*} So we can see that all elements of $B_e(y)$ are included in $I_a^c$ for any value of $y$, so $I_a^c$ is open. A similar argument (with $e=y-b$) shows that $I_b^c$ is open as well. Thus, since the union of two open sets is also open, $I^c$ is open. Since complement of $I$ is open, $I$ is a closed set.
I suggest you don't use the notation $I_a^c$ for the set $\{y\in\mathbb R^1\mid y<a\}$, just use a letter like A or B, because the letter $c$ indicates that this set is the complement of $I_a$ which is not defined. The same for $I_b^c$. After that, everything else is fine.