(New page: Claim: For all <math>a \in \mathbb{R}, A_{0} = \{ x \in \mathbb{R} : g(x) > a \}</math> is open relative to <math>\mathbb{R}</math>. Proof: Suppose <math>\exists x \in A_{0}</math>, fix ...) |
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Proof: Suppose <math>\exists x \in A_{0}</math>, fix this x. Then <math>g(x) > a</math>, and | Proof: Suppose <math>\exists x \in A_{0}</math>, fix this x. Then <math>g(x) > a</math>, and | ||
<math>\exists</math> a set <math>E \subseteq (x,x+1)</math> such that <math>f(y) > a</math> on E, and <math>|E| > 0</math>. | <math>\exists</math> a set <math>E \subseteq (x,x+1)</math> such that <math>f(y) > a</math> on E, and <math>|E| > 0</math>. | ||
| − | Suppose | + | Suppose <math>|E| = \gamma > 0</math>. |
I want to show that there is a small ball of | I want to show that there is a small ball of | ||
| − | radius <math>\delta</math> in <math> mathbb{R}</math> such that this small open ball is also in <math>A_{0}</math>. | + | radius <math>\delta</math> in <math> \mathbb{R}</math> such that this small open ball is also in <math>A_{0}</math>. |
Choose <math>\delta > 0</math> like this: take <math>\delta = \gamma/2</math>. | Choose <math>\delta > 0</math> like this: take <math>\delta = \gamma/2</math>. | ||
Then, <math>\forall y \in \mathbb{R}</math> such that <math>|x - y| < \delta</math>, we have | Then, <math>\forall y \in \mathbb{R}</math> such that <math>|x - y| < \delta</math>, we have | ||
<math>| (y,y+1) \bigcap E | \geq \delta = \gamma/2 > 0</math>. Therefore, <math>g(y) \geq g(x) > a.</math> | <math>| (y,y+1) \bigcap E | \geq \delta = \gamma/2 > 0</math>. Therefore, <math>g(y) \geq g(x) > a.</math> | ||
| − | And thus, for all <math>y \in mathbb{R}</math> such that <math>|x-y| < \delta, y \in A_{0}</math> as well! | + | And thus, for all <math>y \in \mathbb{R}</math> such that <math>|x-y| < \delta, y \in A_{0}</math> as well! |
So, $A_{0}$ is open relative to $R$, and Theorem (4.14) on page 56 tells | So, $A_{0}$ is open relative to $R$, and Theorem (4.14) on page 56 tells | ||
us that g is lsc at x=0 (i.e. if a = g(0)). | us that g is lsc at x=0 (i.e. if a = g(0)). | ||
Latest revision as of 09:44, 29 July 2008
Claim: For all $ a \in \mathbb{R}, A_{0} = \{ x \in \mathbb{R} : g(x) > a \} $ is open relative to $ \mathbb{R} $.
Proof: Suppose $ \exists x \in A_{0} $, fix this x. Then $ g(x) > a $, and
$ \exists $ a set $ E \subseteq (x,x+1) $ such that $ f(y) > a $ on E, and $ |E| > 0 $. Suppose $ |E| = \gamma > 0 $.
I want to show that there is a small ball of
radius $ \delta $ in $ \mathbb{R} $ such that this small open ball is also in $ A_{0} $.
Choose $ \delta > 0 $ like this: take $ \delta = \gamma/2 $.
Then, $ \forall y \in \mathbb{R} $ such that $ |x - y| < \delta $, we have
$ | (y,y+1) \bigcap E | \geq \delta = \gamma/2 > 0 $. Therefore, $ g(y) \geq g(x) > a. $
And thus, for all $ y \in \mathbb{R} $ such that $ |x-y| < \delta, y \in A_{0} $ as well!
So, $A_{0}$ is open relative to $R$, and Theorem (4.14) on page 56 tells
us that g is lsc at x=0 (i.e. if a = g(0)).
