Continuous functions are closed

Question

I am wondering whether every continuous real valued functions is closed? Apparently it seems that it is really that way. Let B be a closed subset of the domain of f. We show that f(B) is closed. If not then there exists a sequence of values $y_n$ in f(B) that converges to a point b, outside f(B). Then there exists a sequence of values $x_n$ in B such that $f(x_n)=y_n$. Since $y_n$ is convergent and f is continuous, so $x_n$ is also convergent to a point a in B (since B is closed) . Since f is continuous, $f(x_n)=y_n$ converges to $f(a)$ which happens to be a point in B, thus contradicting the fact that $y_n$ converges to a point $b$ outside f(B). Is my thinking correct?

Answer 1

Then there exists a sequence of values $x_n$ in $B$ such that $f(x_n)=y_n$. Since $y_n$ is convergent and $f$ is continuous, so $x_n$ is also convergent to a point a in $B$ (since $B$ is closed).

Not quite. It's true that if $x_n$ converges then $f(x_n) = y_n$ also converges, but the converse does not hold. For example, if $f$ is the constant map $0$, then $y_n = 0$ is the only option. But, we can make literally any sequence $x_n$, and it will satisfy $f(x_n) = y_n$. So, think of a sequence that's as non-convergent as you can make it, use it for $x_n$, and it will be a counterexample to this logic.

That said, constant functions are indeed closed. If you want a counterexample, consider $$f(x) = \frac{1}{1 + x^2}.$$ Then $\Bbb{R}$ is closed, and $f(\Bbb{R}) = (0, 1]$, which is neither open nor closed. In particular, if we take any $y_n \to 0$, then inevitably we must have $|x_n| \to \infty$, and therefore $x_n$ cannot be convergent.