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    Hi,

    Would anyone be able to explain how to do this question? Name:  question.PNG
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    There is a solution posted which claims that f is continuous at a value a if and only if a is not an integer, however I don't understand how to reach that conclusion or how to justify it.

    It would be really helpful if someone could also explain how to approach this type of question in general (i.e. working with functions that have continuous and discontinuous regions).

    Thanks,
    Dan
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    (Original post by DanKeitley)
    Hi,

    Would anyone be able to explain how to do this question? Name:  question.PNG
Views: 187
Size:  12.9 KB

    There is a solution posted which claims that f is continuous at a value a if and only if a is not an integer, however I don't understand how to reach that conclusion or how to justify it.
    Let A_x=\{k \in \mathbb{Z} : k \le x\}

    Then e.g.

    A_{2.1}=\{k \in \mathbb{Z} : k \le 2.1\} = \{ \dots, 0, 1, 2 \}
    f(2.1) = \sup A_{2.1} = 2

    A_{1.9}=\{k \in \mathbb{Z} : k \le 1.9\} = \{ \dots, -1, 0, 1 \}
    f(1.9) = \sup A_{1.9} = 1

    So there seems to be a bit of a jump in the graph near 2. It's useful to draw the graph of the function so you can see the discontinuities.

    Now to formalise this, I guess you want to investigate what happens with, say:

    A_{2+\epsilon}=\{k \in \mathbb{Z} : k \le 2+\epsilon\}
    A_{2-\epsilon}=\{k \in \mathbb{Z} : k \le 2-\epsilon\}

    as \epsilon \rightarrow 0, or something along those lines, with some arbitrary n replacing 2.
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    (Original post by atsruser)
    Let A_x=\{k \in \mathbb{Z} : k \le x\}

    Then e.g.

    A_{2.1}=\{k \in \mathbb{Z} : k \le 2.1\} = \{ \dots, 0, 1, 2 \}
    f(2.1) = \sup A_{2.1} = 2

    A_{1.9}=\{k \in \mathbb{Z} : k \le 1.9\} = \{ \dots, -1, 0, 1 \}
    f(1.9) = \sup A_{1.9} = 1

    So there seems to be a bit of a jump in the graph near 2. It's useful to draw the graph of the function so you can see the discontinuities.

    Now to formalise this, I guess you want to investigate what happens with, say:

    A_{2+\epsilon}=\{k \in \mathbb{Z} : k \le 2+\epsilon\}
    A_{2-\epsilon}=\{k \in \mathbb{Z} : k \le 2-\epsilon\}

    as \epsilon \rightarrow 0, or something along those lines, with some arbitrary n replacing 2.
    You're right, I should have drawn the graph! The solution given takes a sequence with  lim_{n \rightarrow \infty} x_n = x and chooses an  \epsilon = min \left\{ x-k,k+1-x \right\} where k<x<k+1 for some integer k. It then shows  lim_{n \rightarrow \infty} f(x_n) = k = f(x) .

    I'm having difficulties with this problem as well which I think is fairly similar. Name:  question.PNG
Views: 43
Size:  16.2 KB.

    I have the solution (see below) but I don't really understand why they've done what they have.

    Name:  answer.PNG
Views: 42
Size:  41.6 KB

    Thanks a lot for your help.
 
 
 
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