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Consider the function $f(x)=[x]$ on the interval $[0,2]$ where $[x]$ denotes the largest integer less than or equal to x.

Is this function continuous? I cant find a reason for it not to be, although im not sure.

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What is the definition of "continuous" that you've been given? – Zev Chonoles Mar 1 '14 at 9:44
See continuity at point 1. – Emin Mar 1 '14 at 9:48
Compute $f(1)$ and $f(3/2)$ and $f(2)$ – Claude Leibovici Mar 1 '14 at 9:50
Suppose $X$ and $Y$ are metric spaces, $E \subset X$, $in E$ and $f$ maps $E$ into $Y$. Then $f$ is said to be continuous at the point $p$ if for every $\epsilon > 0$ there exist a $\delta > 0$ such that: $$d_{Y}(f(x),f(p)) < \epsilon$$ for all points $x \in E$ such that $d_{X}(x,p) < \delta$ Does this matter? Can a function be continuous with one definition and discontinuous with another? – Sodan Mar 1 '14 at 9:52
@George: Continuity is a consistent notion. However, you'll probably want people answering your question to use the definition that you're familiar with. – Zev Chonoles Mar 1 '14 at 9:56
up vote 4 down vote accepted

The function $\lfloor x \rfloor$ is not continuous at interior points that are integers. In this case it at $x=1$. It is also discontinuous at $x=2$

If you are looking for a formal argument here is one.

If a function $f(x)$ is continuous at $x=a$ then for every $\varepsilon>0$ there exist an $\delta>0$ such that $$|x-a|<\delta \Rightarrow |f(x)-f(a)| < \varepsilon $$

So if the function is continuous at $x=1$ then for $\varepsilon=\frac12$ there should be a $\delta$. If there is a delta at points such that $1-\delta<x<1+\delta$ the inequality $|f(x)-1| < \frac12 \, \, \star$ should be satisfied.

Now our job is reduced to show that whatever $\delta$ we choose there is a point for which $\star$ do not hold.

So if we choose a $\delta < 1$ the point $0<1-\frac\delta2<1$ and $f(x)=0$ at all such regions implying that $|f(x)-1|=|0-1|=1>\frac12$.

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The domain of $f$ was given as $[0,2]$. Every answer and comment has (naturally?) focused on the discontinuity at $x=1$ (which is clearly enough), but also the point $x=2$ belongs to the domain and is a point of discontinuity. But no "problem" exists at $x=0$. The asker can check his understanding of these facts. – Jeppe Stig Nielsen Mar 1 '14 at 20:13

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You can surely see that the function is not continuous at integral points. To illustrate it compute $\lim_{x\rightarrow 1^+}[x]$ and $\lim_{x\rightarrow 1^-}[x]$ show that the limit does not exist at $x = 1$. Hence the function cannot be continuous.

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It is not continous.

To see that, consider for example that $f(1) = 1$, but $f(1 - \epsilon) = 0$ for any $\epsilon>0$. There is a 'jump' from $f=0$ to $f=1$ without any values inbetween, regardless of the size of $\epsilon$.

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Hint: Proof the negation of $$ (\forall \epsilon >0)(\exists \delta >0) \Big( (\forall x)(|x-1|<\delta \implies |f(x)-f(1)|<\epsilon)\Big), $$ that is $$ (\exists \epsilon >0)(\forall \delta >0) \Big( (\exists x)(|x-1|<\delta \mbox{ and } |f(x)-f(1)|>\epsilon)\Big) $$

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If you're familiar with the properties of continuous functions, another way to see it is that $f^{-1}(1)=[0,1)$, which is not a closed subset of $[0,2]$.

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Another characterisation of a function $f$ being continous at an argument value $x$ is, that it has a value $f(x)$ there which is equal to its limit value there. This requires an existing limit value.

That last requirement is not valid for your function at $x = 1$:

If you arrive from the left, the limit at $x = 1$ is $0$, $$ \lim_{x \to 1 \atop x < 1} \left[ x \right] = \lim_{x \to 1 \atop 0 < x < 1} \left[ x \right] = \lim_{x \to 1 \atop 0 < x < 1} 0 = 0 $$

but if you arrive from the right, the limit value is $1$,

$$ \lim_{x \to 1 \atop x > 1} \left[ x \right] = \lim_{x \to 1 \atop 2 > x > 1} \left[ x \right] = \lim_{x \to 1 \atop 2 > x > 1} 1 = 1 $$

So the limit is not defined at $x = 1$, but $f$ is defined there. It is not continous there.

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