Proof of coercivity and compactness of a continuous function using contradiction?

Question

The goal here is to prove the following statement:

If the set $\{x|f(x) \le \alpha\}$ is compact, then the function $f$ is coercive.

I saw a so-called answer to this proof, but it does not seem convincing to me. It says

assume each of the sets $\{x|f(x) \le \alpha\}$ is bounded and let $\{x^v\} \subset \mathbb{R}^n$ be such that $||x^v|| \rightarrow > \infty$. Let us suppose that there exists a subsequence of the integers $J\subset \mathbb{N}$ such that the set $\{f(x^v)\}_J$ is bounded above. Then there exists $\alpha \in \mathbb{R}^n$ such that $\{x^v\}_J \subset \{x|f(x)\le \alpha \}$. But this cannot be the case since each of the sets $\{x|f(x)\le\alpha\}$ is bounded while every subsequence of the sequence $\{x^v\}$ is unbounded by definition. Therefore, the set $\{f(x^v)\}_J$ cannot be bounded, and so the sequence $\{f(x^v)\}$ contains no bounded subsequence, i.e. $f(x^v)\rightarrow\infty$.

I do not understand this answer, and specifically this sentence in the answer: “But this cannot be the case since each of the sets $\{x|f(x)\le\alpha\}$ is bounded while every subsequence of the sequence $\{x^v\}$ is unbounded by definition.”

If we are claiming that this sentence is a contradiction, then in that case the very first sentence of this answer, we have already reached the contradiction! The very first sentence reads: “assume each of the sets $\{x|f(x) \le \alpha\}$ is bounded and let $\{x^v\} \subset \mathbb{R}^n$ be such that $||x^v|| \rightarrow \infty$”. Which, as far as i can tell, is the same as what the proof is saying "cannot be the case". Is that true?

I do not understand how this proof is proceeding to show a contradiction, when to me it has already shown it in the first sentence just by assuming a contradiction. Can someone helps me understands this please?

Answer 1

I think you're confusing the sequences $x^v$, which is in $\mathbb{R}^n$ and whose norms tend to $\infty$, and $f(x^v)$, which is in $\mathbb{R}$ and may or may not tend to $\infty$ itself.

That said, I think the proof is a little bit tricky to follow, so let me see if I can write the same idea a little clearer.

We wish to show that $f$ is coercive, that is, given a sequence $(x_k) \in \mathbb{R}^n$ with $\|x_k\| \to \infty$, we must necessarily have $f(x_k) \to \infty$ in $\mathbb{R}$.

Suppose, for the sake of contradiction, that $(x_k) \in \mathbb{R}^n$ such that $\|x_k\| \to \infty$, but $f(x_k) \not\to \infty$. Then there exists some subsequence $x_{k_m}$ such that $f(x_{k_m})$ (not $x_{k_m}$!) is bounded above. Let $\alpha \in \mathbb{R}$ be such a bound, i.e. so that $f(x_{k_m}) \le \alpha$ for all $m$. That is, $$x_{k_m} \in \lbrace x \in \mathbb{R}^n : f(x) \le \alpha \rbrace.$$ Since this set is compact, some convergent subsequence $x_{k_{m_l}}$ exists in the above set. Since $x_{k_{m_l}}$ is convergent, it is also bounded, meaning that $\|x_{k_{m_l}}\| < \beta$ for some $\beta \in \mathbb{R}$. But $\|x_k\| \to \infty$, which means that no subsequence can be bounded, so we have a contradiction.

EDIT: Question 1: A real sequence that does not tend to positive $\infty$ always has a subsequence that is bounded above (in fact, tending to $\infty$ is equivalent to having no subsequence that is bounded above). I am applying this to the sequence $f(x_k)$. To prove this, we can go back to the $M$-$N$ definition of of convergence to $\infty$.

Suppose $a_n$ is a real sequence. By definition, we say $a_n \to \infty$ if, for all $M \in \mathbb{R}$, there exists some $N \in \mathbb{N}$ such that $$n \ge N \implies a_n \ge M.$$ That is, $a_n$ becomes larger than any bound, and eventually stays larger.

If we take the negation of this, $a_n \not\to \infty$ means that, there exists some $M \in \mathbb{R}$ with the property that, for all $N \in \mathbb{N}$, there exists some $n \ge N$ such that $a_n < M$.

If we take $\alpha$ to be this value of $M$, then using $N = 1$, there exists an $n_1 \ge 1$ such that $a_{n_1} < \alpha$. We can do this again with $N = n_1 + 1$. This gives us $n_2 \ge n_1 + 1$ such that $a_{n_2} < \alpha$. Recursively, we continue building this sequence so that $n_{k+1} \le n_k + 1$, but still $a_{n_k} < \alpha$. That is, we've built a bounded subsequence.

Like I said before, I used this on $f(x_k)$. This gave us a bounded above subsequence $f(x_{k_m}) \le \alpha$.

Question 2: Convergence (to something finite) always implies boundedness; there is no need for any kind of compactness. Suppose $x_n \to L$ and pick your favourite value of $\varepsilon > 0$ (I'll take $1$, because I'm boring). By definition of convergence, there exists some $N$ such that $$n \ge N \implies \|x_n - L\| < 1 \implies |\|x_n\| - \|L\|| < 1 \implies \|x_n\| < 1 + \|L\|,$$ by the reverse triangle inequality. This only covers $x_N, x_{N+1}, x_{N+2}, \ldots$. However, the rest is a finite set of points, which is bounded, so we have, $$\|x_n\| \le \max \lbrace \|x_1\|, \|x_2\|, \ldots, \|x_{N-1}\|, 1 + \|L\| \rbrace$$ for all $n$.

Question 3: That's too general. The question says that the lower level sets of the function are compact, which I definitely need in the proof. I'm assuming that the sequence in the codomain $f(x_k)$ does not tend to $\infty$. Then I apply the result from question 1.

The result from question 1 gives me a bounded subsequence. I then take this back to the domain. Boundedness tells me that it belongs to one of these compact lower level sets, so we apply the definition of compactness to get a subsequence in the domain that converges.

Finally, I apply the result from question 2: the convergent subsequence is bounded. This can't happen, since the sequence in the domain is supposed to tend (in norm) to $\infty$.