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    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
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    (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  p/q &lt; a . If we add 1 to it, we have  \dfrac{p+q}{q} &gt; a . But by our assumption of p/q is smallest fraction less than a,  \dfrac{p+1}{q} &gt;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 .
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    (Original post by EnglishMuon)
    Im not sure why that would be continuous as imagine we have chosen the largest fraction  p/q &lt; a . If we add 1 to it, we have  \dfrac{p+q}{q} &gt; a . But by our assumption of p/q is smallest fraction less than a,  \dfrac{p+1}{q} &gt;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

    its question 8 here for reference https://www0.maths.ox.ac.uk/system/f...22/Metric1.pdf
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    (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?
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    (Original post by EnglishMuon)
    hmm strange. do you have a link to the original question?
    yeah edited the post above
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    Let x \in \mathbb{Q}.
    Suppose that x &lt; a, then f(x) = x. Then since f is continuous, for all \varepsilon &gt; 0, there exists \delta &gt; 0 such that for all c \in \mathbb{Q} with |x - c| &lt; \delta, we have that |x - c| &lt; \varepsilon.
    Let \eta := \min(\varepsilon, a - x) &gt; 0, then for all c \in \mathbb{Q} with |x - c| &lt; \eta,
     |f(x) - f(c)| = |x - c| &lt; \varepsilon,
    hence f is continuous for x &lt; a

    A similar argument gives f continuous for x &gt; a. The key here is to note a \notin \mathbb{Q}, so the discontinuity that would occur in the reals never occurs in the rationals. The c \in \mathbb{Q} 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.
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    (Original post by Alex:)
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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.
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    (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 [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.
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    (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 [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.
    edit: nvm I was misreading the thing and thinking the domain was the range hence my confusion!
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