Is this function continuous on transcendental number This question is motivated from Thomae's function continuity at irrationals together with the fact that transcendental numbers are dense in real numbers.
Let $$f(x) = \begin{cases}1 &,  \text{x is algebraic}\\ 0 &,  \text{ x is transcendental}. \end{cases}$$
I "think" $f(x)$ is continuous over transcendental numbers but I can't prove it.however I may be wrong. Thanks in advance.
 A: No, this function lacks the essential property of Thomae's function, which is that it is close to zero on many rationals.  In particular, for any $\varepsilon  > 0$, Thomae's function is smaller than $\varepsilon$ in an open ball around every irrational number.  Your function would be continuous at a transcendental number $x$ only if there were an open interval $(x−\delta,x+\delta)$ containing no algebraic numbers. This never happens, as the algebraic numbers are dense in $\mathbb{R}$.
On the other hand, it should be possible to modify your function to obtain one with the desired property.  For instance, each algebraic number is a root of some $n$-th order polynomial with integer coefficients $a_0,a_1,\ldots, a_n$.  Define the size of a polynomial to be $\max\{n, |a_0|, |a_1|,\ldots, |a_n|\} \ge 1$, and define the size $S(x)$ of an algebraic number to be the smallest size of any polynomial with $x$ as a root.  Then the function
$$
\cases{0 & \text{if $x$ is transcendental}\\ 1/S(x) & \text{if $x$ is algebraic}}
$$
is continuous at every transcendental number and discontinuous at every algebraic number.
