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Stewart - Calculus

In the red box, is that really 'a stronger' rather than 'an additional' ?

I mean, if simply-connected implies open connected, then why do we have open simply-connected in Theorem 6?

Otherwise, what is meant by stronger? I thought that meant simply-connected implies open connected.

Note: My knowledge on topology or whatever you call it is mainly limited to 1D.

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When we assume that the region is simply connected, you're right that we're just making an additional assumption about the region. The author calls this condition 'stronger' because it's harder to satisfy. There are plenty of open, connected regions which fail to be simply connected.

We do, however, have to include the word open in Theorem 6. Not all simply connected regions are open.

211792
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    Thanks! Counterexamples please? – BCLC Oct 16 '16 at 14:28
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    Consider the annulus ${(x,y)|1<x^2+y^2<2}\subset\mathbb{R}^2$. This region is open and connected, but not simply connected. On the other hand, the closed disk ${(x,y)|x^2+y^2\leq 1}\subset\mathbb{R}^2$ is simply connected, but not open. Theorem 6 only applies when the region is both open and simply connected. – 211792 Oct 16 '16 at 14:30
  • Thanks Austin Christian! ^-^ – BCLC Oct 16 '16 at 14:32
  • Actually wait I think I made a mistake. Open and simple is stronger than open? :p – BCLC Sep 26 '17 at 09:06