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?

  • $\begingroup$ 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 $[1, \omega]$? $\endgroup$ – 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.


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