Let $f: \mathbb{R} \to \mathbb{R}$ be continuous. Show the set

$$A:= \{\alpha \in \mathbb{R}: \exists \{x_n\}_{n \in \mathbb{N}} \subset \mathbb{R} \, \, \text{ with} \, \, \lim_{n \to \infty} f(x_n) = \alpha \}$$

is connected.

Proof idea:

Since this is a subset of $\mathbb{R}$. This is equivalent to showing that, given $\alpha_1, \alpha_2 \in A$, if $\alpha_1 < z < \alpha_2$ then $z \in A$. Let $\epsilon < \min\{d(\alpha_1,z), d(\alpha_2,z)\}$. Let $\{x_n\} \to \alpha_1$ and $\{x'_n\} \to \alpha_2$. There exists an $N \in \mathbb{N}$ so that, for $f((x_N,x'_N)) \subset (\alpha_1 -\epsilon, \alpha_2-\epsilon)$.
Since $f$ is continuous, the IVT gives that there exists $z_1 \in (x_N,x'_N)$ with $f(z_1) = z$. Iteratively, this gives a sequence $\{z_n\}$ with $f(z_n) = z_0$ and $\lim \limits_{n \to \infty} f(z_n) = z$. Ergo $z \in A$ so $A$ is connected.

Any issues with this proof? Any other ways to attack this problem?


$A$ is the closer of $f(\Bbb R)$ i.e $A=\overline{f(\Bbb R)}$. As $\Bbb R$ is connected and $f$ continuous, then $f(\Bbb R)$ is connected, hence the closer is connected.

  • $\begingroup$ Much cleaner, thank you. $\endgroup$ – Anthony Peter Sep 14 '15 at 19:40

Different approach: start as you did, with the notion of $z$ as the target.

Let $x_i$ and $y_i$ be sequences such that that $f(x_i)$ and $f(y_i)$ approach $\alpha_1$ and $\alpha_2$ respectively.

Observe that for large enough $n$,e.g., $n > N$, we have $d(f(x_n),\alpha_1) < d(z, \alpha_1)$, and similarly for $d(f(y_n), \alpha_2) < d(z, \alpha_2)$.

For such $n$, we have $f(x_n) < z < f(y_n)$.

By IVT there's a number $q \in [x_n, y_n]$ with $f(q) = z$.

Now pick $$z_1, z_2, \ldots = q$$

and you're done.


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