How to prove that f is discontinuous at one point at least Let $f: [0,1] \rightarrow \{0,1\}$
with $f(0) = 0$
and  $f(1) = 1$
$\{0,1\}$ has the discrete topology on it.
How to prove that $f$  has at least one discontinuity. One cannot apply the intermediate value theorem because there isn't any value between $f(0) = 0$ and $f(1) = 1$.
Is it possible to prove it without using connectedness?
Many thanks in advance.
 A: It is clear that $f$ is onto $\{0,1\}$ by the condition that $f(0) = 0$ and $f(1) = 1$. If $f$ is continuous on $[0,1]$ that is connected, we would have that $f([0,1]) = \{0,1\}$ is connected. But this is a contradiction because I can write $\{0,1\} = \{0\} \cup \{1\}$, both of which are open in the subspace topology on $\{0,1\}$. Hence $f$ is discontinuous at at least one point in $[0,1]$.
By the way, if you put the trivial topology on $\{0,1\}$ then $f$ is continuous on $[0,1]$.
A: Suppose, for the sake of contradiction, that $f$ is continuous. By definition, the inverse of open sets is open. 
If we give $\{0,1\}$ the discrete topology, then the inverses of $\{0\}$ and $\{1\}$ are both open. By hypothesis, they are nonempty. They union to $[0,1]$, so we have produced a separation of $[0,1]$. If we assume $[0,1]$ has the usual topology, then it is connected (this is a standard fact). But a connected set cannot have a separation, so we get a contradiction, and $f$ cannot be continuous.
A: HINT: Think of $f$ as a function from $[0,1]$ to $\Bbb R$ and use the intermediate value theorem to conclude that it cannot be continuous.
A: In the following a point $\xi$ of discontinuity is produced by means of binary search, and without appeal to "higher" theorems:
Put $I_0:=[0,1]$, and for each $n\geq1$ let $I_n$ be the left or the right half of $I_{n-1}$, where each time the choice is made such that $f$ takes different values at the endpoints of $I_n$. There is a point $\xi$ which is contained in all $I_n$. The function $f$ is not continuous at $\xi$, whatever the value $f(\xi)$.
A: This proof uses least upper-bound property which I am guessing might be equivalent to assuming connectedness of real numbers, since for $\mathbf{Q}$ I would not be able to assume the largest of the $\delta$ to exist. 
Assume that the function is continuous, then consider an arbitrary point $x_{0} \in \left[0,1\right]$. Then, for every $\epsilon > 0$, there must exist a $\delta > 0$ such that $0<d\left(x,x_{0}\right)<\delta \implies \left(f(x),f(x_{0})\right) < \epsilon$ 
Let $x_{0} = 0$, then for any $\epsilon$, there exists a $\delta_{0}>0$, hence in the interval $[0,\delta_{0})$, $f(x)=0$. Now we can say that for no $\delta > d_{0}$ is the condition satisfied, because if it is, then we can take the new $\delta$ as $\delta_{0}$ and go on increasing till we get $\delta_{0}=1$. 
Now for any interval around $\delta_{0}$, it can be shown that the condition for continuity is not satisfied. 
