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There is a theorem that says that if $f$ is a continuous map, then its graph is homeomorphic to the domain. The converse is false. I'd like to find the simplest counterexample possible. I've found this

$f:\mathbb{R}\longrightarrow \{0,1\}$,

where $\mathbb R$ is endowed with the topology generated by the sets $\{1\},\{2\},\{3\},...$ and $f(x)=1$ for $x=0,1,2,...$ and $f(x)=0$ otherwise.

Is there a simpler example?

edit: In my example, $\{0,1\}$ has discrete topology.

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What do you mean by domain here? An open connected set? –  Paul Jan 21 '12 at 14:51
    
No! By "the domain" I meant the domain of $f$. en.wikipedia.org/wiki/Function_%28mathematics%29#Definition –  Bartek Jan 21 '12 at 15:05
    
Oh I see. Got it. How stupid I am! I keep thinking of domain in the set of point-set topology. –  Paul Jan 21 '12 at 15:06
    
Does anyone know a reference for the result that the graph of a continuous function is homeomorphic to its domain? –  Michael Greinecker Jan 21 '12 at 16:46
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I don’t think that your example works. The topology on $\mathbb{R}$ gives it $\omega$ isolated points, and the only nbhd of any other point is all of $\mathbb{R}$. The graph of $f$ has $\omega$ isolated points, but $\langle 0,f(0)\rangle$ is a non-isolated point that has a proper subset of the graph as a nbhd, and the isolated points of the graph are not dense in the graph. –  Brian M. Scott Jan 21 '12 at 18:52

2 Answers 2

up vote 4 down vote accepted

Here's one that seems easy to visualize.

Take $X = [0,1) \times \mathbb{N}$ with its usual topology; i.e. $X$ is a countable disjoint union of half-open "sticks". Define $f : X \to \mathbb{N}$ by $$f(x,n) = \begin{cases} 1, & n=1, x < 1/2 \\ 2, & n=1, x \ge 1/2 \\ n+1, & n > 1.\end{cases}$$ Then a homeomorphism from the graph of $f$ to $X$ is given by $$F((x,n),k) = \begin{cases} (2x,1), & k=1 \\ (2x-1, 2), & k=2 \\ (x,k), & k > 2. \end{cases}$$ (Hopefully I have written it correctly.)

The idea is that the graph looks just the same, except that the first stick has been broken in half. By stretching the half-pieces we get two full-size sticks. But there were already infinitely many sticks, so adding one more doesn't produce an "extra" if we re-index.

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Nate’s is nicer, but here’s an example based on the proposer’s original idea.

Topologize $\mathbb{Z}$ as follows: each even integer is an isolated point, and if $n$ is odd, a set $V$ is a nbhd of $n$ iff $V\supseteq\mathbb{Z}\setminus 4\mathbb{Z}$. Call the resulting space $X$. Let $A=\{4n+1:n\in\mathbb{Z}\}$, give $D=\{0,1\}$ the discrete topology, and let $f:X\to D$ be the indicator (characteristic) function of $A$. Clearly $f$ is not continuous, since $A=f^{-1}[\{1\}]$ is not open in $X$.

Now let $G$ be the graph of $f$. The set of isolated points of $G$ is

$$I=\Big(A\times\{1\}\Big)\cup\Big(2\mathbb{Z}\times\{0\}\Big)\;.$$

If $p\in G\setminus I$, then $p=\langle 4n+3,0\rangle$ for some $n\in\mathbb{Z}$, and a set $V\subseteq G$ is a nbhd of $p$ in $G$ iff $V\supseteq \Big(\mathbb{Z}\setminus \big(A\cup 4\mathbb{Z}\big)\Big)\times\{0\}$. From this it’s not hard to check that the function

$$h(n)=\begin{cases} \left\langle \frac{n}2,0\right\rangle,&\text{if }n\equiv 0\pmod 8\\\\ \left\langle\frac{n}2-1,1\right\rangle,&\text{if }n\equiv 4\pmod 8\\\\ \langle n,0\rangle,&\text{if }n\equiv 2\pmod 4\\\\ \langle 2n+1,0\rangle,&\text{if }n\equiv 1\pmod 2 \end{cases}$$

is a homeomorphism of $X$ onto $G$: the first two clauses match up the isolated points that don’t have to belong to nbhds of the non-isolated points; the third matches up the other isolated points; and the fourth matches up the non-isolated points.

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