The set of points of discontinuity of the greatest integer function [x], is

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Let f(x) = [x] be the greatest integer function. Let k be any integer.

Then,

$f(x)=[x]=\left\{\begin{array}{cl}
k-1, & \text { if } k-1 \leq x<k \\
k, & \text { if } k \leq x<k+1
\end{array}\right.$                          [By def.]

Now, 

(LHL at x = k) = $\lim\limits_{x \rightarrow k^{-}} f(x)=\lim\limits_{h \rightarrow 0} f(k-h)=\lim\limits_{h \rightarrow 0}[k-h]$

⇒ (LHL at x = k) = $\lim\limits_{h \rightarrow 0}(k-1)=(k-1)$

and, (RHL at x = k) = $\lim\limits_{x \rightarrow k^{+}} f(x)=\lim\limits_{h \rightarrow 0} f(k+h)=\lim\limits_{h \rightarrow 0}[k+h]$

⇒ (RHL at x = k) = $\lim\limits_{h \rightarrow 0} k \quad[∵ k \leq k+h<k+1 ∴ [k+h]=k]$

⇒ (RHL at x = k) = k

∴  $\lim\limits_{x \rightarrow k^{-}} f(x) \neq \lim\limits_{x \rightarrow k^{+}} f(x)$

So, f(x) is not continuous at x = k

Since k is an arbitrary integer. Therefore, f(x) is not continuous at integer points.

Let a be any real number other than an integer. Then, there exists an integer k such that k - 1 < a < k

Now,

(LHL at x = a) = $\lim\limits_{x \rightarrow a^{-}} f(x)=\lim\limits_{h \rightarrow 0} f(a-h)=\lim\limits_{h \rightarrow 0}[a-h]$

⇒ (LHL at x = a) = $\lim\limits_{x \rightarrow 0} k-1 \quad\left[\begin{array}{l}∵ k-1<a-h<k \\ ∴ [a-h]=k-1\end{array}\right]$

⇒ (LHL at x = a) = k - 1

⇒ (RHL at x = a) = $\lim\limits_{x \rightarrow a^{+}} f(x)=\lim\limits_{h \rightarrow 0} f(a+h)=\lim\limits_{h \rightarrow 0}[a+h]$

⇒ (RHL at x = a) = $\lim\limits_{h \rightarrow 0} k-1 \quad\left[\begin{array}{l}∵ k-1<a+h<k \\ ∴ [a+h]=k-1\end{array}\right]$

⇒ (RHL at x = a) = (k - 1)

and, f(a) = k - 1         [∵ k - 1 < k < a      ∴ [a] = k - 1]

Thus, $\lim\limits_{x-a^{-}} f(x)=\lim\limits_{x \rightarrow a^{+}} f(x)=f(a)$

So, f(x) is continuous at x = a. Since a is an arbitrary real number, other than an integer, therefore f(x) is continuous at all real points except integer points.