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Question: Give the set of points of discontinuity of $f(x)= \lfloor x \rfloor$.

Answer: When $x=0.9999, y=0$ but when $x=1, y=1$, so for a very small change in the value of $x$, the change in the value of $y$ is big and this happens at all the integers (positive as well as negative), therefore the set of all the integers is the answer. Am I correct?

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    Yes, you are correct.2011-10-08
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    Formally, you could use the definition of continuity, say, with epsilon = 1/22011-10-08
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    Yes Its true for $x\in \mathbb{Z}$2011-10-08
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    Rigorously speaking, the values of f at x=0.9999 and at x=1 do not prove nor disprove the continuity of f at x=1. Hence I assume that 0.9999 stands for *any value less than 1 and close to 1*... in which case your idea is correct but cannot be called a proof. Yet.2011-10-08
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    You have a clear *geometric* idea of what's going on. Depending on the type of course, you may be expected to write it up semi-formally (as $x$ approaches an integer $n$ from the left, $\lfloor x\rfloor$ does not approach $\lfloor n \rfloor$) or formally ($\epsilon$-$\delta$). Don't forget that you also need to show that your list of points of discontinuity is complete, that is, that if $a$ is *not* an integer, then the function $\lfloor x\rfloor$ is continuous at $a$.2011-10-08
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    If you like epsilons and deltas, you can say that at any integer, if you set $\varepsilon=1/2$, then there is no $\delta$ so small that if the distance from $x$ to the integer in question is $<\delta$, then the distance from $f(x)$ to the value of the function at the integer in question is $<\epsilon$. For some of those values of $x$, the distance will be $1$. But at non-integers, no matter how small $\varepsilon$ is, it is enough to make $\delta$ small enough so that the neighborhood of radius $\delta$ contains no integers.2011-10-08

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