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Is $A=\{(x,y): x^2+y^2=1\}$ is connected in $ℝ^2$?

From its graph, I would conclude that it's not path connected.

Royi
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Sandy
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  • Isn't that just $S^1$, which can be identified with $[0, 2\pi) \subseteq \mathbb R$? My topology is a bit rusty but that's what comes to mind. –  Mar 14 '18 at 15:46
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    What is your definition of connected? And what does the graph look like? – Servaes Mar 14 '18 at 15:47
  • It's connected, the easiest way to see this is to use what @tilper have given you, it is the continuous image of $[0,2\pi)$ which is also connected. – Jürgen Sukumaran Mar 14 '18 at 15:51
  • @Servaes the graph is boundary of unit circle – Sandy Mar 14 '18 at 15:52
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    @Sandy which is clearly connected in the plain english sense of the word. So that makes me wonder what your definition is, and why you think it isn't path connected. – Servaes Mar 14 '18 at 15:56
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    @Sandy perhaps you're confusing connected with convex? – Arnaud D. Mar 14 '18 at 16:03

1 Answers1

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$$ A = \left\{(x,y) : x^2 + y^2 = 1 \right\} = \left\{(x,y) : x = \cos \theta, y = \sin \theta, 0 \leq \theta < 2\pi\right\} $$

The interval $\Theta = \left[0,2\pi \right)$ is connected, since $A$ is image of continuous map defined on a connected set then it is also connected (as proved here)

user8469759
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