I have seen multiple proofs online that use continuity to prove that the set is connected. But how will we go about doing this without the use of mapping? Is there any way we can do this by using the definition of connectedness (not a union of non-empty disjoint open sets)?


2 Answers 2


Every path is an arc and hence lies in the circle. Therefore the circle is path-connected and hence connected.

  • $\begingroup$ Sorry, but what does "path-connected" mean? $\endgroup$ Mar 8, 2020 at 14:04
  • $\begingroup$ That still uses continuity arguments. $\endgroup$ Mar 8, 2020 at 14:13

Maybe you can, but I know of no such proof. You need to show that connectedness is preserved by continuous maps anyway (and that proof is quite trivial), to show it is a topological property, and for many applications (intermediate value theorem among others). There is no reason to avoid using it.

  • $\begingroup$ We hadn't done continuity in class yet (we're following Rudin's book) and was wondering if there was a way of doing a proof without it. Thanks though! $\endgroup$ Mar 8, 2020 at 14:04
  • $\begingroup$ @overloadedMathNerd In Rudin's book he doesn't ask in the exercises to show connectedness of the circle; he sneaks in a sort-of-continuity argument in an exercise on convex sets (but the circle isn't convex). So in the excercises you're not following Rudin, then? $\endgroup$ Mar 8, 2020 at 14:23
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    $\begingroup$ we're basically following the book in terms of lectures and notes. But questions are usually novel and made by the Professor himself (or taken from some other source). The question wasn't exactly about the unit circle. It just asked for a counter-example to the idea that intersection of connected sets is connected. I thought I'd use the unit circle as part of my answer. $\endgroup$ Mar 8, 2020 at 15:36

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