Sequential compactness in $\mathbb{R}$

Question

Well known result:
Suppose $f:\mathbb{R}\to \mathbb{R}$ is continuous and let $K$ be a compact set. Then, $f(K)$ is compact.

I can prove this using the definition of compactness (finding a finite subcover), but I was trying to see if it would work using sequential compactness since the two are equivalent in $\mathbb{R}$. If it does work, the proof would go something like this:

Let $\{y_n\}$ be a sequence in $f(K)$. Then $\{x_n\}$ defined by $x_n:= f^{-1}(y_n)$ is a sequence in $K$.

Since $K$ is compact, it is sequentially compact. So, there's a subsequence $\{x_{n_j}\} = \{f^{-1}(y_{n_j})\}$ with $f^{-1}(y_{n_j}) \to x \in K$ as $j \to \infty$.

Since $f$ is continuous, we have that $f(x_{n_j})=y_{n_j} \to f(x) \in f(K)$.

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I feel like there might be a problem here because we don't know $f$ is one-to-one. What do you think?

Answer 1

It is better to write ; for $n\in \mathbb{N}$, ($y_n\in f(K)$) there exists $x_n\in K$ such that $y_n=f(x_n)$, to avoid $f^{-1}$...

Answer 2

This is a correct proof. It relies on the fact that $f(f^{-1}(y_{n_j}))=y_{n_j}$ which is true as $f$ is a function. Being one-to-one would guarantees that $f^{-1}(f(y))=y$ but that's not needed here.

Answer 3

Let $ y_n = f(x_n)$ be a sequence in $f(K)$. Then we have a sequence $(x_n)$ in $K$. Since $K$ is compact, there exists a subsequence $(x_{n_k})$ of $(x_n)$ such that $$\lim_{k \to \infty} x_{n_k} = x \in K.$$ But this means that $(y_n)$ has a subsequence $y_{n_k} = f(x_{n_k})$ such that $$\lim_{k \to \infty} y_{n_k} = \lim_{k \to \infty} f(x_{n_k}) = f \left( \lim_{k\to \infty} x_{n_k} \right) = f(x) = y \in f(K).$$ Since we took an arbitrary sequence in $f(K)$ and showed it has a subsequence that converges in $f(K)$ we have proved that $f(K)$ is compact.