Given a quasi-ordered set $(Q,\leq)$ the interval topology on $Q$ is generated by $$\{Q\setminus\downarrow x : x\in Q\} \cup \{Q\setminus\uparrow x : x\in Q\},$$ where $\downarrow x = \{y\in Q: y\leq x\}$ and $\uparrow x = \{y\in Q: y\geq x\}$.

Let $\mathbb{N}^\mathbb{N}$ denote the set of all functions $f:\mathbb{N}\to\mathbb{N}$ and define a quasi-ordering relation $\leq^*$ on $\mathbb{N}^\mathbb{N}$ by $$f\leq^* g \text{ iff } \exists k\in \mathbb{N} \text{ such that } f(n) \leq g(n) \text{ for all } n\geq k.$$

Endow $\mathbb{N}$ with the discrete topology and let $\tau$ be the product topology on $\mathbb{N}^\mathbb{N}$. Let $\tau_i(\mathbb{N}^\mathbb{N})$ be the interval topology of $(\mathbb{N}^\mathbb{N}, \leq^*)$.

The identity map $\text{id}: (\mathbb{N}^\mathbb{N},\tau)\to (\mathbb{N}^\mathbb{N},\tau_i(\mathbb{N}^\mathbb{N}))$ is continuous, which gives rise to a natural question:

Is there a continuous surjection $f: (\mathbb{N}^\mathbb{N},\tau_i(\mathbb{N}^\mathbb{N}))\to (\mathbb{N}^\mathbb{N},\tau)$?


In view of the answer to this question, the answer is no, for $\tau_i$ is connected, but $\tau$ is not.


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