Let $E = \{(x,y) \in \mathbb{R}^2 \mid y=x^2 \}$. Show that $(E,d_E)$ is a complete metric space.

Question

Let $E = \{(x,y) \in \mathbb{R}^2 \mid y=x^2 \}$. Show that $(E,d_E)$ is a complete metric space. Here $d_E$ is just the restriction of $d_2$ to $E$.

I'm trying to show this using the preimage definition. I have that $$E=\{(x,y)\mid f(x,y)=0\}$$ where $f(x,y)=y-x^2$. Note that $f$ is continous so if $E$ is the preimage of a closed set then $E$ is closed and since $\mathbb{R}^2$ is complete and $E \subset \mathbb{R}^n$ this would imply that $E$ is also complete. I'm not sure I entirely understand the concept of preimage. It's quite clear that it's just the set of elements that map to the image of $f$, but how do I find it? I'm tempted so say that $$f^{-1}(\{0\}) = E$$ but this would mean that only $0$ gets mapped to $0$ which isn't true since if $x=y=1$ then $f(x,y)=0$ so the preimage would be $f^{-1}(\{0,1\})$?

Answer 1

Given any function $g:X\to Y$ and $A\subseteq Y$ we have that

$$g^{-1}(A)=\{x\in X\ |\ g(x)\in A\}$$

This is simply the definition.

And so in your case $f^{-1}(\{0\})=E$ pretty much by the definition.

but this would mean that only 0 gets mapped to 0

That doesn't even make sense, since $0\not\in\mathbb{R}^2$ and $f^{-1}(\{0\})\subseteq\mathbb{R}^2$. Note that $(x,y)\in f^{-1}(\{0\})$ means that $f(x,y)=0$. Its equivalent.

if $x=y=1$ then $f(x,y)=0$

Yes, and thus for $x=y=1$ we have that $(x,y)=(1,1)$ is an element of $f^{-1}(\{0\})$. And at the same time an element of $E$.

Answer 2

We define $F(x,y)=x^2-y$. So $F^{-1}(\{0\})=\{(x,y):x^2=y\}=E$, and since $F$ is continuous as you mentioned, and ${0}$ is closed in $\Bbb{R}$ we get that $E$ is closed in $\Bbb{R^2}$ (as the preimage of a closed set under a continuous function). So $E\subseteq\Bbb{R^2}$ is a closed subset in a complete metric space and therefore $(E,d_E)$ is complete.

Answer 3

About your concern regarding the preimage notation: $$E = f^{-1}(\{0\})$$ is really just another way of writing the definition of $E$: $$(x,y) \in f^{-1}(\{0\}) \iff f(x,y)\in \{0\} \iff f(x,y) = 0.$$

Answer 4

That's true that $f$ is continuous, and $E = f^{-1}(0)$, and since point zero is closed in $\mathbb R$ then by continuity of $f$ set $E$ is closed.You misunderstand the symbol $f^{-1}$, it's not and inverse map, we define the preimage of set $X \subset \mathbb R$ the same way$$f^{-1}(X) = \{(x, y) \in \mathbb R | f(x,y) \in X\}$$