Continuous function?

Question

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.

Answer 1

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.

Answer 2

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 $|f(x)-1|=|0-1|=1>\frac12$.

Answer 3

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 $\epsilon$.

Answer 4

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 $[0,2]$.

Answer 5

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 $$ (\exists \epsilon >0)(\forall \delta >0) \Big( (\exists x)(|x-1|<\delta \mbox{ and } |f(x)-f(1)|>\epsilon)\Big) $$

Answer 6

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.