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Let $S^1$ denote the circle. I want to write $S^1=U\cup V$ with $U$ and $V$ being open. I read that we should take $U$ and $V$ to be open arcs slightly bigger than semicircles. I want to understand why the semicircle is not open but when the arcs $U$ and $V$ are taken slightly bigger than semicircles they become open.

I know that the upper semicircle is $$S_+=\{(x,y)\in \mathbb R^2\;|\;y=\sqrt{1-x^2}\}$$ and the lower semicircld is $$S_-=\{(x,y)\in \mathbb R^2\;|\;y=-\sqrt{1-x^2}\}$$ but i don't know whey they are not open.

2 Answers 2

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I want to write $S^1=U\cup V$ with $U$ and $V$ being open.

Open in what?

If you want $U$ and $V$ to be open in $\mathbb R^2$, then that's not possible, because a finite union of open sets is open, and $S^1$ is not open in $\mathbb R^2$.

If you want $U$ and $V$ to be open in the subspace topology of $S^1$, then you can take $U=V=S^1$... Or, if you want two "a little more than semicircles," you can take any small value of $\epsilon > 0$ and take $$U=S^1\cap \{(x,y)| y>-\varepsilon\}\\ V=S^1\cap \{(x,y)| y<+\varepsilon\}\\ $$

Also, note that $U\cap V\neq \emptyset$ in this case. This must (in the standard topology) always be the case because $S^1$ is connected.

Edit:

The semicircle isn't open because the points $(1, 0)$ and $(-1, 0)$ do not have any open neighborhood that are subsets of either semicircle.

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    $U$ and $V$ open in $S^1$. I want to know why the semicircles do not verify this. – 2017-02-09
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    is it because the upper semicircle is $S^1\cap \{(x,y)| y\ge 0\}$ and $ \{(x,y)| y\ge 0\}$ is closed in $\mathbb R^2$ so $S^1\cap \{(x,y)| y\ge 0\}$ is closed in $S^1$ ? – 2017-02-09
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    @palio See the edit. – 2017-02-09
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    @palio Technically, just because a set is closed doesn't mean it can't also be open. – 2017-02-09
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i don't know whey they are not open.

In general if $f:X\to Y$ is continous with $Y$ Hausdorff then the graph

$$Gr(f)=\{(x,y)\in X\times Y\ |\ y=f(x)\}$$ is a closed subspace of $X\times Y$. In your case $S_{+}$ (and analogously $S_{-}$) is a graph of $f_{+}:[-1,1]\to\mathbb{R}$, $f_+(x)=\sqrt{1-x^2}$. Note that it is closed in $[-1,1]\times \mathbb{R}$. Obviously the intersection $Gr(f_{+})\cap S^1$ is closed in $S^1$ by definition.

Now can $S_+$ be open? Assume it is. Then it is both closed and open and thus $S^1\backslash S_{+}$ is closed and open. In particular those two sets would give a partition of $S^1$ into open sets (the technicality here is that both are nonempty). But $S^1$ is connected (being an image of $[0, 1]$). This is a contradiction.