How do I know $x \in \bigcap\limits_{n = 1}^\infty f^{-1}([n,\infty))$?

Question

Suppose $X$ is a compact metric space and $f: X \rightarrow \mathbb{R}$ is a function for which $f^{-1}([t,\infty))$ is closed for any real $t$. Then $f$ achieves its maximum value on $X$.

I am first trying to prove that that $f$ is bounded above. Suppose, for the sake of contradiction, that $f$ is not bounded above. So for all $M \in \Bbb{R}$, there is a $x \in X$ such that $f(x) > M$. So there exists a sequence $\{x_n\}$ in $X$ such that for all $n \in \Bbb{N}$, we have $f(x_n) > n$. So $x_n \in f^{-1}([n,\infty))$ for all $n \in \Bbb{N}$. Since $f^{-1}([n,\infty)) \subseteq X$ are closed for all $n \in \Bbb{N}$, $f^{-1}([n,\infty))$ are compact for all $n \in \Bbb{N}$. Further, since each $f^{-1}([n,\infty))$ is nonempty, and $f^{-1}([n + 1,\infty)) \subseteq f^{-1}([n,\infty))$ for all $n \in \Bbb{N}$, $\bigcap\limits_{n = 1}^\infty f^{-1}([n,\infty))$ is not empty, and it is also closed.

I am trying to show now that $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ that converges to some $x \in \bigcap\limits_{n = 1}^\infty f^{-1}([n,\infty))$, and reach a contradiction since $$\bigcap\limits_{n = 1}^\infty f^{-1}([n,\infty)) = f^{-1}(\bigcap\limits_{n = 1}^\infty [n,\infty))$$ is empty (since $\bigcap\limits_{n = 1}^\infty [n,\infty)$ is empty).

My issue is, how do I know $x \in \bigcap\limits_{n = 1}^\infty f^{-1}([n,\infty))$? I know its in $X$ since it is compact, but I don't know if its in $x \in \bigcap\limits_{n = 1}^\infty f^{-1}([n,\infty))$.

Help, please.

Answer 1

You cannot know that, since it is not true. Suppose that $x\in\bigcap_{n=1}^\infty f^{-1}\bigl([n,\infty)\bigr)$. Then $(\forall n\in\mathbb{N}):f(x)\geqslant n$. But no real number is greater than or equal to every natural number.

Answer 2

Here is a slightly more general statement:

Claim. Let $K_1 \supseteq K_2 \supseteq \cdots $ be non-empty closed subsets of a metric space $X$. If $x_n \in K_n$ for each $n$, then every subsequential limit of $(x_n)$ lies in $K_{\infty} := \bigcap_{n=1}^{\infty} K_n$.

• For instance, we may apply this to $K_n = f^{-1}([n, \infty))$.

• This claim does not involve compactness. As such, $K_{\infty}$ can be empty. In particular, if $K_{\infty} = \varnothing$ then $(x_n)$ has no subsequential limit in $X$.

Proof. Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$ and write $x$ for its limit in $X$. Then for each fixed $N$ and for each $k \geq N$, we have $x_{n_k} \in K_{n_k} \subseteq K_{n_N}$. So, by the closedness of $K_{n_N}$, we have $x \in K_{n_N}$. But since this is true for every $N$, we must have $x \in \bigcap_{N=1}^{\infty} K_{n_N} = K_{\infty}$.

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As for the last equality, we have

Lemma. If $A_1 \supseteq A_2 \supseteq \cdots$ is a sequence of sets, then for any subsequence $(A_{n_k})$ we have $\bigcap_{k=1}^{\infty} A_{n_k} = \bigcap_{n=1}^{\infty} A_n$.

Proof. To show that two sets are equal, we may try to show that one contains the other.

First, assume that $x \in \bigcap_{k=1}^{\infty} A_{n_k}$. This means that $x \in A_{n_k}$ for each $k$. Now, for each $n$, there exists $k$ such that $n_k \geq n$, and so, $x \in A_{n_k} \subseteq A_n$. Since this is true for all $n$, we have $x \in \bigcap_{n=1}^{\infty} A_n$, and so, $\bigcap_{k=1}^{\infty} A_{n_k} \subseteq \bigcap_{n=1}^{\infty} A_n$.

The other direction $\bigcap_{n=1}^{\infty} A_n \subseteq \bigcap_{k=1}^{\infty} A_{n_k}$ is almost obvious. Indeed, if $x \in \bigcap_{n=1}^{\infty} A_n$, then $x \in A_n$ for any $n$, and so, $x \in A_{n_k}$ for any $k$.