# a sequence defined by continuous function on $\mathbb{N} \cup\{ \infty\}$

Let $$f$$ be a continuous complex-valued function on $$\mathbb{N} \cup\{ \infty\}$$, one-point compactification of $$\mathbb{N}$$.

Let $$\{a_n \}$$ be a sequence defined by $$a_n =f(n)$$ for each $$n \in \mathbb{N}$$. Then what can we say about the sequence $$\{a_n \}$$?

First, I read from a different post that $$\mathbb{N} \cup\{ \infty\}$$ is homeomorphic to the compact and bounded set $$\{ \frac{1}{n}: n \in \mathbb{N} \} \cup \{0\}$$, then can we say that the sequence must be bounded (because it is the image of compact set under a continuous function) and if so, can we say that the sequence is convergent to $$0$$?

• What if $f(x)=1$?
– Joe
Commented May 6, 2021 at 0:36

Continuous functions send limits to limits, and since in $$\mathbb{N} \cup \{\infty\}$$, $$\lim_{n \to \infty} n = \infty$$, it is the case that $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} f(n) = f(\lim_{n \to \infty} n) = f(\infty).$$ Also as you note, $$\mathbb{N} \cup \{\infty\}$$ is compact, and the image of compact sets under continuous maps are compact (and thus bounded in $$\mathbb{C}$$), so the sequence is bounded. Convergent sequences are also bounded, so the above explanation is a bit overkill.