How do I prove that this set is closed in $\mathbb{R}_+ \times \mathbb{R}$?

Question

The set is $$\{(x, y) \in \mathbb{R}_+ \times \mathbb{R} : y \in \{\pm \sqrt{x}\}\}$$

I have tried to see if the complement is open but this example is tricky and I would like to ask for some help on how to do that. Thanks.

Answer 1

For the set $S=\{(x,y)\in{\bf{R}}_{+}\times{\bf{R}}: y=\sqrt{x}\}=\{(x,\sqrt{x}): x\in{\bf{R}}_{+}\}$, if $(x_{n},\sqrt{x_{n}})\in S$ is such that $(x_{n},\sqrt{x_{n}})\rightarrow(x,y)$, then $x_{n}\rightarrow x$, and the continuity of $\sqrt{\cdot}$ gives $\sqrt{x_{n}}\rightarrow\sqrt{x}$, but $\sqrt{x_{n}}\rightarrow y$, so $y=\sqrt{x}$, so $(x,y)=(x,\sqrt{x})\in S$.

Answer 2

Prescribe $f:\mathbb R_+\times\mathbb R\to\mathbb R$ by: $$\langle x,y\rangle\mapsto x-y^2$$

It can be shown to be continuous so that $f^{-1}(\{0\})$ as preimage of a closed set will be closed.

Now observe that $f^{-1}(\{0\})$ is the set mentioned in your question.