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Prove that $V(x,y) \rightarrow (y,-x)$ is continuous

My attempt:

Let $c \in \mathbb{R}^2$ and let $x$ exist in the delta neighborhood of $c$ in $\mathbb{R}^2$ ($x\in V_\delta(c) \cap \mathbb{R}^2$). For any $\epsilon >0$ there exists a $\delta >0$ such that

$|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$. In other words:

$|x-c|=|(x_1,x_2)-(c_1,c_2)|= \sqrt{(x_1 - c_1)^2+(x_2-c_2)^2}<\delta$ implies $|f(x)-f(c)|=|(x_2,-x_1)-(c_2,c_1)|=\sqrt{(x_2-c_2)+(x_1-c_1)}<\epsilon$

Not sure where to go from here or if my approach is correct

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    You're missing the squares under the square root in the last expression. Notice that $\|x-c\|=\|f(x)-f(c)\|$ because $(x_1-c_1)^2+(x_2-c_2)^2 = (x_2-c_2)^2+(x_1-c_1)^2$. Hence, just pick $\delta < \varepsilon$. – 2017-02-20

2 Answers 2

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Projections are continuous; negation is continuous $\mathbb{R} \to \mathbb{R}$; taking a Cartesian product is continuous; and compositions of continuous functions are continuous.

So $$[(x,y) \mapsto y] \times ([u \mapsto -u] \circ [(x,y) \mapsto x])$$ is continuous.

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Given $\epsilon > 0$, let $\delta = \epsilon$.

Then $|x-c|=|(x_1,x_2)-(c_1,c_2)|= \sqrt{(x_1 - c_1)^2+(x_2-c_2)^2}<\delta$ implies $|f(x)-f(c)|=|(x_2,-x_1)-(c_2,-c_1)|=\sqrt{(x_2-c_2)^2+(x_1-c_1)^2}<\epsilon$

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    In the second line you must have $(c_2,-c_1)$, not $(c_2,c_1)$. – 2017-02-20