Homeomorphism of $K$ and $K\cup \{0\}$

Question

Given $K = \{\frac{1}{n} | n > 0\}$ can we construct a homeomorphism between $K$ and $K \cup 0$ given the discrete, codiscrete, and standard topologies.

We can see that $f: K\cup\{0\} \to K$ given by $0 \mapsto \frac{1}{1}$ and $\frac{1}{n} \mapsto \frac{1}{n+1}$. This is a bijection since we can see $f(K\cup\{0\}) = K$ and if $f(\frac{1}{n_1}) = f(\frac{1}{n_2})$, then $\frac{1}{n_1+1} = \frac{1}{n_2+1} \implies n_1 = n_2$. So this means we just have to check continuity of the inverse, $g$.

For the discrete topology, all open sets are every possibly collection. So we see that $g^{-1}(\{\frac{1}{n_1}\} \cup ... \cup \{\frac{1}{n_i}\}) = g^{-1}(\frac{1}{n_1}) \cup ... \cup g^{-1}(\frac{1}{n_i}) \in \mathcal{T}_{disc}$, and $g^{-1}(0) = 1 \in \mathcal{T}_{disc}$. So all open sets in the image are open in the preimage.

As for the codiscrete topology, we see that $g^{-1}(\emptyset) = \emptyset$ and $g^{-1}(K) = g^{-1}(\{\frac{1}{n} | n > 0\}) = K \cup \{0\} \in \mathcal{T}_{codisc}$. So it is a homeomorphism.

Finally, for $\mathcal{T}_{st}$, we see that for any open in $K \cup \{0\}$, it will either be entire set, finite $\frac{1}{n_i}$ with or without $0$, or all $\frac{1}{n}, n>0$. So just checking $g^{-1}(\frac{1}{n_1} \cup ... \cup \frac{1}{n_i}) = \{\frac{1}{n_i -1}| i<n\} = \{\frac{1}{n_i -1}| i<n\} \cap (\frac{1}{n_i} - \epsilon,2) \in \mathcal{T}_{st}$ where we choose $\epsilon$ to be smaller than the difference between the consecutive terms. If, it's the whole set, then we just choose $K \cap (0,2) \in \mathcal{T}_{st}$.

Is there some detail that I'm missing?

Answer 1

Let's collect some facts:

• Any map $f:(X,T_\text{discrete})\to (Y,T)$ is continuous.
• Any map $f:(X,T)\to (Y,T_{\text{codiscrete}})$ is continuous.
• If $f:X\to Y$ is continuous and $X$ is compact, so is $Y$.

Can you conclude now?