# A function converges if and only if it can be extended to a continuous map

Let $$X$$ be any topological space and $$f:\mathbb{Z}_{+}\longrightarrow X$$ a sequence in $$X$$.

Consider $$\mathbb{Z}_{+}\subset[1,\omega]=\mathbb{Z}_{+}\cup\{\omega\}$$, where $$\omega$$ is the first infinite ordinal.

We topologies $$[1,\omega]$$ with the order topology.

Show that the sequence $$\big\{f(n)\big\}_{n\in\mathbb{Z}_{+}}$$ converges to $$\alpha\in X$$ if and only if $$f$$ extends to a continuous map $$F:[1,\omega]\longrightarrow X$$ by declaring $$F(\omega)=\alpha$$.

By now, I don't quite have an idea for this question. I know that since $$\mathbb{Z}_{+}$$ is the discrete topology, then $$f$$ is automatically continuous.

Any hints about how I should consider this question?

• Where have you started? Can you prove that if $f$ converges to $\alpha$ then $F$ is continuous? How about the converse? Do you know the definition of contintuity? Can you describe the open sets of $[1, \omega]$? – Mees de Vries Oct 3 '18 at 8:36

If we define $$F$$ the way you mentioned, then, for every open set $$A\subset X$$ containing $$\alpha$$, $$F^{-1}(A)$$ contains $$\omega$$ and every $$n$$ when $$n$$ is large enough. So, $$F^{-1}(A)$$ is an open subset of $$\mathbb{Z}_+\cup\{\omega\}$$. This proves that $$F$$ is continuous.
On the other hand, if $$\alpha$$ is not a limit of the sequence $$\bigl(f(n)\bigr)_{n\in\mathbb N}$$, then, for every open set $$A$$ such that $$\alpha\in A$$, $$F^{-1}(A)$$ will not contain infinitely many $$n$$'s (but it contains $$\omega$$), and therefore it is not an open subset of $$\mathbb{Z}_+\cup\{\omega\}$$. This proves that $$F$$ is not continuous.