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Let $f\colon \mathbb R\rightarrow \mathbb R$ be a continuous function. Define $G = \{(x, f(x)) : x \in \mathbb R\} \subseteq \mathbb R^2$. Pick out the true statements:

a. $G$ is closed in $\mathbb R^2$.

b. $G$ is open in $\mathbb R^2$.

c. $G$ is connected in $\mathbb R^2$.

I think c is correct since $f$ is continuous but no idea about a and b.

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up vote 7 down vote accepted

Write $G=\{(x,y),y=f(x)\}$ and define $g\colon \Bbb R^2\to \Bbb R$, $g(x,y)=y-f(x)$. It's a continuous map and $G=g^{-1}(\{0\})$.

$G$ can't be open: $(0,f(0))\in G$, but $(0,f(0)+r)\notin G$ for $r\neq 0$.

$G$ is connected as the range of the connected set $\Bbb R$ by the continuous map $F\colon \Bbb R\to \Bbb R^2$, $F(x)=(x,f(x))$.

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+1: Nice solution for (a)! I'd not thought to approach it that way. – Cameron Buie Sep 5 '12 at 12:25
Dumb question: Why does this show (a) again? (I'm sure its correct, I'm just slow at understanding it... :p) – Legendre Sep 5 '12 at 12:35
$\{0\}$ is closed in $\Bbb R$, and $g^{-1}(F)$ is closed if $F$ is closed and $g$ continuous. – Davide Giraudo Sep 5 '12 at 12:38
+1: Doh! Thanks and nice solution! – Legendre Sep 5 '12 at 12:49

a is true: $(x_n,f(x_n))\to(a,b)\implies x_n\to a,~f(x_n)\to b.$ Since $f$ is continuous on $\mathbb R,~x_n\to a\implies f(x_n)\to f(a).$ Since limit of a sequence is unique for a metric space $b=f(a)\implies (a,b)\in G.$

b is false: $f\equiv 0\implies G=\mathbb R\times \{0\}.$

c is true: Continuous image of a connected set is connected. Now since the projections $\mathbb R\to\mathbb R:x\mapsto x,~\mathbb R\to\mathbb R:x\mapsto f(x)$ are continuous so is $\mathbb R\to\mathbb R^2:x\to(x,f(x)).$

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