1

Let $h : \mathbb{R} \to \mathbb{R}$ be a continious function.Define $\Gamma(h) =\{ (a,b) \in \mathbb{R}\times \mathbb{R} : b=h(a)\}$

My question: Is $\Gamma(h) $ closed in $\mathbb{R}^2?$

My attempt : I think No .Take $r: \mathbb{R}^2 \to \mathbb{R} $ and defined by $$r(a,b)=b-h(a)$$

since $ b-h(a)=0 \implies (a,b)= r^{-1}(\{0\})$

Thus $\Gamma(h) =r^{-1}(\{0\})$

$(a,b)$ is open in $\mathbb{R}^2$

This implies $\Gamma(h)$ is open in $\mathbb{R}^2$

Asinomás
  • 107,565
jasmine
  • 15,021
  • @Yorch i edited – jasmine Aug 10 '21 at 22:29
  • 2
    Where you write $b-h(a)=0\Rightarrow (a,b)=r^{-1}({0})$, you should have $b-h(a)=0\Rightarrow (a,b)\in r^{-1}({0})$. I think this is what leads to the statement "$(a,b)$ is open in $\mathbb{R}^2$", which doesn't really make sense since $(a,b)$ is an element of $\mathbb{R}^2$ and not a subset. – Julian Rosen Aug 10 '21 at 22:32
  • 2
    You should have written ${(a, b)}$ rather than $(a, b)$ and ${(a, b)}$ is a closed subset of $\Bbb{R}^2$ for any $a$ and $b$. Simple sketching of functions like $x \mapsto 1$ or $x \mapsto 2x$ should have convinced you that the graph of a continuous function is likely to be closed. – Rob Arthan Aug 10 '21 at 22:38
  • 1
    You also can prove directly using the $\varepsilon$-$\delta$ definition that $\Bbb R^2 \setminus \Gamma(h)$ is open in $\Bbb R^2$ when $h$ is continuous. – Robert Shore Aug 10 '21 at 22:54

1 Answers1

1

We have $\Gamma(h)$ is the preimage of $\{0\}$ under the continuous map $r:\mathbb R\times \mathbb R \rightarrow \mathbb R$ given by $r(a,b) = b - h(a)$. Since $\{0\}$ is closed in $\mathbb R$ we conclude $\Gamma(h)$ is closed in $\mathbb R \times \mathbb R$, as desired.

If $\{0\}$ was open then we could conclude that $\Gamma(h)$ is open, but $\{0\}$ is not open .

Asinomás
  • 107,565