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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.

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Hint: consider $\sup\{a\in[0,1]\mid f(a)=0\}$. – Arturo Magidin Jul 21 '12 at 4:00
@LuoKaisa Why do you want to prove it now without connectedness? – user38268 Jul 21 '12 at 14:19
@BenjaLim I only wanted to have a direct proof without having to resort to other theorems. The hint of Arturo led me in that direction. Thanks. – Luo Kaisa Jul 21 '12 at 19:50
up vote 4 down vote accepted

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 $f(\xi)$.

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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 $[0,1]$.

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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.

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Your proof is essentially the same as mine..... – user38268 Jul 21 '12 at 4:08
Indeed. If I had seen your answer, I probably wouldn't have posted it. – Potato Jul 21 '12 at 4:11
The proofs are pretty much the same, and any proofs are going to use connectedness. But, if you are looking for a difference, this proof doesn't explicitly use the result that continuous images of connected sets is connected (as small of a result as that is). So there is that. – Francis Adams Jul 21 '12 at 4:14
@FrancisAdams The problem is not true if you put the trivial topology on $\{0,1\}$ – user38268 Jul 21 '12 at 4:15
@BenjaLim It also isn't true if you put the discrete topology on $[0,1]$, but I would assume that isn't what is being asked. – Francis Adams Jul 21 '12 at 4:46

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.

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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.

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