Prove a function is continuous

Let a be an irrational number. Show that f: Q to Q is continuous, where
f(x) = x if x<a
x+1 if x>a

Not really sure here, my idea is to maybe consider subsets S of Q, show f^-1 (S) is always closed (or open), maybe using the fact that an arbitrary union of closed sets is closed (not sure this works though as Q is infinite, or does the fact that Q is countably infinite make it work?)

Or should I be looking to use the epsilon- delta definition (if so does it matter which metric I equip Q with)?

Things with rationals and irrationals are not my fav, any help appreciated xx

@Gregorius

Im not sure why that would be continuous as imagine we have chosen the largest fraction $p/q < a$. If we add 1 to it, we have $\dfrac{p+q}{q} > a$. But by our assumption of p/q is smallest fraction less than a, $\dfrac{p+1}{q} >a$ suggesting there are other rationals in the interval $[\dfrac{p+1}{q} ,\dfrac{p+q}{q}]$ which we have missed out iff $q \not= 1$.

Yeah i thought that adding 1 would be too big of a jump when i first read the question but it says its continuous

hmm strange. do you have a link to the original question?

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I think I understand your argument but how does that f(x) = x+1 ..if x>a affect it rather than f(x)=x?. Are we saying that the 'gap' caused by the x+1 and x doesnt effect the continuity? I understand that each segment for x<a and x>a is continuous separately but not sure why it is all together.

I realise it's quite unintuitive, but the gap that exists at a is at an irrational point. Just like how if we're proving continuity on $[a,b] \subset \mathbb{R}$ and we don't have to consider anything below a or above b; we're proving continuity on $\mathbb{Q} \subset \mathbb{R}$, so we don't consider any irrational numbers in the domain.

The function is well defined and we can approach a as close as we wish, but we never reach it, and hence never reach the discontinuity that is in the irrationals.