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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
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Welcome to math.SE. Since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, statements phrased in the imperative ("Give", "Prove") when asking for help are considered rude by many here; please consider rewriting. –  Arturo Magidin Jul 21 '12 at 4:02
    
@ArturoMagidin I apologize for my previous post. I rewrote it. Thanks for the hint. –  Luo Kaisa Jul 21 '12 at 14:18
    
@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
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5 Answers 5

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
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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
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@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
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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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