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How do you show the complement of these subsets are open?

  1. $S^1=\{(x_1,x_2)\in\Bbb R^2\mid x^2_1+x^2_2=1\}$
  2. $B^c_1=\{(x_1,x_2)\in\Bbb R^2\mid x^2_1+x^2_2\ge1\}$
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Try to prove the corresponding versions of the statements for $\mathbb{R}^1$ first. How do you prove that $S^1 = \{ x \in \mathbb{R} : x^2 = 1 \}$ and $\{ x \in \mathbb{R} : x^2 \geq 1 \}$ are closed (i.e. their complements are open)? You pick a point $x$ in the complement and find a sufficiently small ball of radius $\epsilon > 0$ centered at $x$ so that the entire ball is contained in the complement... Hope that helps get you started. – Michael Joyce Mar 1 '12 at 0:34
@kahen: Edited. Please go easy on folks who are new to this site. – Aryabhata Mar 1 '12 at 0:35
Looks like $x21$ is a typo. I suppose you mean $x12$. – Patrick Mar 1 '12 at 0:36
@Patrick: I guess he copy-pasted from a HTML or PDF exercise sheet, accidentally missing the superscript/subscript formatting; the original must have been produced by software that puts the superscript first on general principles. – Henning Makholm Mar 1 '12 at 0:42

I'll do the second problem:

$\color{darkgreen}{B_1}$ is the unit ball of radius 1 centered at the origin. Note that it contains its boundary. Any point at distance 1 from the origin is on $\color{darkgreen}{B_1}$.

Now, to prove that $B_1^C$ is open, you have to show the following:

for any point $\color{maroon}{x_0}$ that is not in $\color{darkgreen}{B_1}$, there is some open ball $\color{maroon}{B_{x_0}(\epsilon)}$, centered at $\color{maroon}{x_0}$, that is disjoint from $\color{darkgreen}{B_1}$. Here, $\color{maroon}{B_{x_0}(\epsilon)} =\{\,z\in\Bbb R^2 \mid d(x_0,z)<\epsilon\,\}$.

So, first pick a point $\color{maroon}{x_0}$ that is not in $\color{darkgreen}{B_1}$. We know then that the distance from $\color{maroon}{x_0}$ to the origin is $r>1$ for some real number $r$.

We need to find an open ball centered at $\color{maroon}{x_0}$ that is disjoint from $\color{darkgreen}{B_1}$. Note that the distance from $\color{maroon}{x_0}$ to the boundary of $\color{darkgreen}{B_1}$ is $r-1>0$ (recall $r>1$). We suspect that, if we let the radius, $\epsilon$, of our ball be less than $r-1$, then the open ball $\color{maroon}{B_{x_0}(\epsilon)}$ will "work".

So, let's take the open ball to be $B_{x_0}({r-1\over2})$.

Now, we have to prove that this open ball works:

Of course, by definition $B_{x_0}({r-1\over2})$ is centered at $\color{maroon}{x_0}$.

Now we have to show that $B_{x_0}({r-1\over2})$ is disjoint from $\color{darkgreen}{B_1}$. To do this, we have to show that if $y\in B_{x_0}({r-1\over2})$, then $y\notin \color{darkgreen}{B_1}$.

So let $y\in B_{x_0}({r-1\over2})$. We have to show that the distance from $y$ to the origin is greater than $1$. Towards this end, using the reverse triangle inequality, and the fact that $y$ being in $B_{x_0}({r-1\over2})$ guarantees that $d(y,x_0)<{r-1\over 2}$: $$ d(y,0)\ge\bigl|\, d(x_0,0)-d(y,x_0)\,\Bigr|> r- {r-1\over 2} ={r+1\over2}>1. $$

And we are done. Any point outside of $B_1$ is contained in an open ball disjoint from $B_1$; thus $B_1^C$ is open.

enter image description here

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Show that the sets are closed. To show that they are closed, observe that (1) is the inverse image of a point by a continuous function, and (2) is the inverse image of a closed subset of the real line by a continuous function.

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