Proving Homeomorphism with Yoneda Lemma: The Unification of epsilon-delta and epsilon-N Formulation

Let us consider the submetric space of $\mathbb{R}$$\mathbb{R}$:

$$S:=\left\{0\right\}\cup \{\frac{1}{n}\mid n\in {\mathbb{N}}_{>0}\}\subset \mathbb{R}.$$

We want to prove that $S\cong {\mathbb{N}}^{\ast }$$S \cong \mathbb{N}^*$ via the Yoneda Lemma.

Recall that in last blog, we defined a functor $F$$F$, which is representable:

$$F\cong {\mathrm{Hom}}_{\mathrm{Haus}}({\mathbb{N}}^{\ast },-).$$

By the Yoneda Lemma, if we can prove that $F\cong {\mathrm{Hom}}_{\mathrm{Haus}}(S,-)$$F \cong \mathrm{Hom}_{\mathrm{Haus}}(S, -)$, then we obtain $S\cong {\mathbb{N}}^{\ast }$$S \cong \mathbb{N}^*$.

Let us do it.

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Proposition

Let $S:=\{0\}\cup \{\frac{1}{n}\mid n\in {\mathbb{N}}_{>0}\}\subset \mathbb{R}$$S := \{ 0 \} \cup \left\{ \frac{1}{n} \mid n \in \mathbb{N}_{>0} \right\} \subset \mathbb{R}$, equipped with the subspace topology inherited from $\mathbb{R}$$\mathbb{R}$. Let $f:S\to X$$f : S \to X$ be a function into a Hausdorff space $X$$X$. Then the following are equivalent:

1. $f(1/n)\to f(0)$$f(1/n) \to f(0)$;

2. $f$$f$ is continuous.

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Proof

Direction (1) $\Rightarrow$$\Rightarrow$ (2): If the sequence converges, then $f$$f$ is continuous

We only need to check continuity at $x=0$$x = 0$, since $\{\frac{1}{n}\mid n\in {\mathbb{N}}_{>0}\}$$\left\{ \frac{1}{n} \mid n \in \mathbb{N}_{>0} \right\}$ is discrete.

Let $U_0$$U_0$ be an open neighborhood of $f(0)$$f(0)$. By the definition of convergence, there exists $N\in \mathbb{N}$$N \in \mathbb{N}$ such that $\mathrm{\forall }n\ge N$$\forall n \ge N$, we have $f(1/n)\in U_0$$f(1/n) \in U_0$.

Hence,

$$f^{-1}(U_0)=\{0\}\cup \{\frac{1}{n}\mid n\ge N\}\cup \{\frac{1}{n}\mid f(1/n)\in U_0\text{ and }n<N\},$$

which is open in $S$$S$.

Direction (2) $\Rightarrow$$\Rightarrow$ (1): If $f$$f$ is continuous, then $f(1/n)\to f(0)$$f(1/n) \to f(0)$

Assume $f$$f$ is continuous. Then, since continuous functions map convergent sequences to convergent sequences, we have

$$f(1/n)\to f(0).◻$$

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Hence, we see that

$${\mathrm{Hom}}_{\mathrm{Haus}}({\mathbb{N}}^{\ast },-)\cong F\cong {\mathrm{Hom}}_{\mathrm{Haus}}(S,-).$$

By the Yoneda Lemma, it follows that ${\mathbb{N}}^{\ast }\cong S$$\mathbb{N}^* \cong S$.

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Remark

Now, we can see that there is no difference between the $ϵ$$\epsilon$–$\delta$$\delta$ and $ϵ$$\epsilon$–$N$$N$ formulations. The $ϵ$$\epsilon$–$N$$N$ language is simply the $ϵ$$\epsilon$–$\delta$$\delta$ definition applied to the point $0\in S$$0 \in S$.