Prove that $A =\{x \in X\mid f(x) \ge 2022\}$ is closed in $X$, by means of sequences. 
Let $f:X\to \mathbb{R}$ be a continuous function, where $(X,d)$ is a metric space, and $\mathbb{R}$ has the usual metric, $|x-y|$. Prove that the set $A=\{x\in X:f(x)\geq 2022\}$ is a closed set in $X$, by means of sequences.

I was thinking of proving that the complement of $A$ is open.
So, I did the following:
Let $0<r<x_0-2022$.
If $x\in B(x,r)\implies x\in A^c$
$$d(x,x_0)=|x-x_0|<r \implies x_0-r<x<r+x_0$$
But
$$\begin{align*}
0<r<x_0-2022
&\implies -r>2022-x_0 \\
&\implies -r-x_0>2022\\
&\implies x>2022\\
&\therefore A^c \text{ is open} \\
&\therefore A \text{ is closed}
\end{align*}$$
But I think I am not using the sequences.
 A: The $2022$ is obviously a red-herring, so I'll forego that detail.

Claim: Let $f : X \to \mathbb{R}$ be continuous, and $c \in \mathbb{R}$ be given. Then $A := \{ x \in X \mid f(x) \ge c \}$ is closed.


*

*Key Facts:

*

*For a metric space $(X,d)$ and $S \subseteq X$, $S$ is closed if and only if: for any convergent sequence $\{x_n\}_{n \in \mathbb{N}}$ with limit $x \in X$, we have $x \in S$. (That is, a subset is closed if convergent sequences in that set converge in that set.)


*Limiting behavior is a topological invariant, in the sense that if $f : X \to Y$ is continuous and $x_n \to x$ in $X$, then $f(x_n) \to f(x)$ in $Y$.




*Sketch of Proof: (I'll let you fill in the details.)

*

*Suppose $\{x_n\}_{n \in \mathbb{N}} \subseteq A$ converges to $x \in X$. (Note: our goal is to show $x \in A$.)

*Find where $\{f(x_n)\}_{n \in \mathbb{N}}$ must lie. (Follows from $A$'s definition.)

*Our goal: use this information to ascertain what value $f(x)$ takes on.

*By way of contradiction, suppose $x \not \in A$. What does this imply for $f(x)$? (Hint: Show that $f(x_n)$ is eventually within $B(f(x);r)$, where $r$ is half the distance between $f(x)$ and $c$, for all $n$ sufficiently large. Why can we claim this? (It's from a definition.) What does this contradict?)

*Hence, $x \in A$ and thus $A$ is closed.




Comments on Your Approach:
Personally, I can't make much out of it, and I don't want to go item-by-item and try to ask questions on each detail when it's largely "what is this" and "why". Some key points I would think about however:

*

*What is $x_0$? What is the $x$ you refer to throughout the proof?

*Why must $r$ lie in that range? I feel like this is unnecessarily restrictive and for no clear reason.

*How do your final implications suddenly invoke $x$?

*How have you shown $A^c$ is open? How does $x > 2022$ make this possible? (This ties back into the $r$ thing, because I think you're trying to show that it contains an open subset?)

In general, your proof could make do with a lot more clarification, and - to be honest - just words in general. Implications and symbols are nice, but they're hard to parse the meaning of. That is why a lot of proofs use ample amounts of English explanations, to explain what is going on and why - not everything need be communicated symbolically. That's half the art of proof-writing: it's not just a chain of implications but also a communication with the reader and attempt to convince them of the validity of your claims.
This would be especially true of schoolwork, since you need to show the person grading you that you know what you're talking about. (In general, be mindful of your audience.)
