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Prove a function is continuous Watch

1. 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
2. (Original post by Gome44)
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 . If we add 1 to it, we have . But by our assumption of p/q is smallest fraction less than a, suggesting there are other rationals in the interval which we have missed out iff .
3. (Original post by EnglishMuon)
Im not sure why that would be continuous as imagine we have chosen the largest fraction . If we add 1 to it, we have . But by our assumption of p/q is smallest fraction less than a, suggesting there are other rationals in the interval which we have missed out iff .
Yeah i thought that adding 1 would be too big of a jump when i first read the question but it says its continuous

its question 8 here for reference https://www0.maths.ox.ac.uk/system/f...22/Metric1.pdf
4. (Original post by Gome44)
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?
5. (Original post by EnglishMuon)
hmm strange. do you have a link to the original question?
yeah edited the post above
6. Let .
Suppose that , then . Then since is continuous, for all , there exists such that for all with , we have that .
Let , then for all with ,
,
hence is continuous for

A similar argument gives continuous for . The key here is to note , so the discontinuity that would occur in the reals never occurs in the rationals. The part of the proof is the single most important thing when proving any sort of continuity, especially when you move away from the real numbers.
7. (Original post by Alex:)
.
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.
8. (Original post by EnglishMuon)
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 and we don't have to consider anything below a or above b; we're proving continuity on , 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.
9. (Original post by Alex:)
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 and we don't have to consider anything below a or above b; we're proving continuity on , 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.
edit: nvm I was misreading the thing and thinking the domain was the range hence my confusion!
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