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Proving a function is continuous

  1. Nov 2, 2007 #1
    Problem statement
    Given f:[1,2]->R defined by f(x) = x^2. Show that this function is continuous

    Problem Solution (my version at least)
    1- It is known sequence {a[n]b[n]} converges to {ab}
    2- Definition of continuous: if every sequence {cn} in f we have f(cn) -> f(c)

    3- Given our domain of 1<= f <=2 assume our sequence an-> a=1 and bn-> b=2
    4- Therefore f(anbn) = f(1*2) = f(a*b) = f(1*2) = 4

    5- Therefore this function is continuous.

    My concern is am I taking a leap of faith in 3,4 which is the essence of my proof.
    Is there a better way to prove this or is this mathematically sufficient.


  2. jcsd
  3. Nov 2, 2007 #2


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    Have you tried or thought about a delta/epsilon proof?
  4. Nov 2, 2007 #3
    Yes... and I couldn't get anywhere with that.
    Infact, I am currently trying to work that out right now -still struggling.

    Is my above reasoning completely off the track?
  5. Nov 2, 2007 #4


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    It's not the most obvious method.

    Where did you stumble in the delta/epsilon proof?
  6. Nov 2, 2007 #5
    This is where I got stuck

    1- problem is to prove that f(x) = x^2 is continuous
    2- Definition of continuous: if every sequence {cn} in f we have f(cn) -> f(c)
    3- Assuming x->a, definition of limit |x^2 - a^2 | < epsilon
    4- x = a + b (b is very small)
    5- Substituting 4 into 3, I get |b^2 + 2ab| < epsilon
    6- I can choose b to be either epsilon/4a or square_root (epsilon/2)

    This is where I crash

    - My problem is what is the next step
    - And how do I go from step 6 to proving the actual problem that x^2 is continuous in [1,2]. Thats why I went to the other proof which I had.

    I agree if I can prove like this it would be more mathematically sound.

    Any pointers on my issues would be appreciated.


  7. Nov 3, 2007 #6


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    Your DEFINITION of "continuous" is in terms of limits of sequences? Then don't use "epsilon-delta" to prove continuity, use properties of sequences.

    If, as appears from your first post, you have already proved: "If {an} converges to a and {bn} converges to b then anbn} converges to ab" then you're home!

    If {xn is any sequence that converges to a, then that property tells you that {xn2}= {xnxn} converges to a2.
    Last edited by a moderator: Nov 3, 2007
  8. Nov 3, 2007 #7
    I thought my first solution was acceptable but enumaelish was saying it is not clear. And I had first approached this problem using 'epsilon' method.

    So in my 1st approach, are steps3, 4 right or should I remove them?
  9. Nov 3, 2007 #8


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    3 and 4 make no sense at all! First, f(anbn) is NOT equal to f(ab). I presume you mean the limit but you cannot just say that "[itex]lim_{n\rightarrow\infty} f(a_n b_n)= f(ab)[/itex]". That's only true if f is continuous at ab and that is what you are trying to prove! Second, "f(1*2)= f(a*b)= f(1*2)" is just silly. It's trivially true since you have defined a= 1 and b= 2 but why say such a thing? Third, you only use the fact that f(x)= x2 when you say f(1*2)= 4. Surely, you need that throughout the proof?

    Finally, IF you had proved the crucial step, that "[itex]lim_{n\rightarrow\infty} f(a_n b_n)= f(ab)[/itex]", you would have proved that f is continuous at x= 2 but only there!

    You need to do what I suggested: prove that, for any x in [1,2], if {[itex]x_n[/itex]} is a sequence in [0,1] converging to x, then {[itex]x_n^2[/itex]} converges to x2. If you have already proved the general theorem, that if {[itex]a_n[/itex]} is a sequence converging to a and {[itex]b_n[/itex]} is a sequence converging to b, then the sequence {[itex]a_nb_n[/itex]} converges to ab, just use that with a and b both equal to x. If you have not already proved it do so! It's more general than proving that {[itex]x_n
    ^2[/itex]} converges to x2 directly and exactly the same work.
    Last edited by a moderator: Nov 3, 2007
  10. Nov 3, 2007 #9
    sequential continuity can be easy to work with, just know the limit laws well. Like Halls of ivy said prove the more general result, that is the limit of a product of convergent sequences is the product of each of the limits of the convergent sequences (if a_n goes to a, b_n goes to b, then a_nb_n goes to ab).

    You can prove an even more general property about continuous functions from here: the product of continuous functions is continuous.

    Then proving rational functions and polynomials are continuous become easy.
  11. Nov 3, 2007 #10
    Hehe, I remember a similar question on my analysis final some years ago. The question was to prove that a product of continuous functions is continuous. The question becomes trivial if you used the sequential criterion for continuity but less so if you want to use a delta-epsilon approach. My professor was tempted to drop a mark as I did not consider whether x_n may have been converging to an isolation point. So the limit of f(x_n)g(x_n) won't even exist. The sequential criterion for continuity only coincides with the delta epsilon one if the limit of the sequence is an accumulation point.

    Do a delta-epsilon proof if you know the delta-epsilon definition of continuity.
  12. Nov 3, 2007 #11


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    Tempted to drop a mark? You mean he didn't??

    Perhaps you just misphrased that but xn can't "converge to an isolation [I would say "isolated"] point"- that pretty much the definition of "isolated point". Perhaps you meant that you didn't consider the case of x being an isolated point of the set.

    But the definitions are equivalent even there: When x is an isolated point, the hypothesis "if {x_n} is a sequence coverging to x" is false and so, by that definition, the function is trivially continuous at x, just as with the "epsilon-delta" definition.
    Last edited by a moderator: Nov 3, 2007
  13. Nov 3, 2007 #12
    Hmm, it's been awhile but sequences can converge to isolated points (think of the discrete metric space)...the sequence just has to eventually become constant. Limits of functions at isolated points don't exist. So if a sequence converges to some isolated point we trivially have f(x_n)->f(x) as x_n is eventually x. However, lim_a->x f(a) is undefined (so it's not f(x)).

    I believe I have a tendency to call isolated points 'isolation points' because I think of isolated points as non accumulation points.
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