Since $\Bbb Q$ is countable, there exists a bijection $f:\Bbb N \to \Bbb Q$. Now consider the same function $f$ with the codomain changed to $\Bbb R$. That is, $f:\Bbb N \to \Bbb R$. My question is, whether the given function is continuous or not(the metric involved is the usual metric).
I'm not able to wrap my head around this thing. I request anyone to shed some light on this matter.
On
I think the heart of the issue is you have not really specified what topology you want $\mathbb{N}$ to have. Lets not think about topologies induced by metrics.
If you pick the discrete topology, every function is continuous. If you pick the other trivial topology, then since the interval $(0,1)$ contains some but not all rational numbers, no bijection $f$ extended in this way will be continuous.
So what topology does $\mathbb{N}$ inherit from "the usual" metric? It will be endowed with the discrete topology, and so any function at all will be continuous, in particular the one you want to consider.
Yes it is, although it is somewhat counterintuitive at the first glance. A moment's further thought can dispel the delusion. Recall the definition of continuity of a real-valued function $f$ defined on a metric space at a point $c$. Take the metric space to be a subset of $\mathbb{R}$ for illustration. The continuity requirement is that for every $\varepsilon > 0$ there is some $\delta > 0$ such that if $|x-c| < \delta$ then $|f(x)-f(c)| < \varepsilon$. Now if $f$ maps $\mathbb{N}$ to $\mathbb{R}$ and if $c \in \mathbb{N}$, then every $0 < \delta < 1$ is such that $|x-c| < \delta$ implies $|f(x) - f(c)| = |f(c) - f(c)| = 0 < \varepsilon$. So by definition $f$ is continuous at $c$. Since $c$ was arbitrary, we proved that $f$ is continuous on $\mathbb{N}$. Since $f$ was arbitrary, we proved that every $\mathbb{N} \to \mathbb{R}$ is continuous.
We may note that this continuity is in some sense "boring".