Help with proving supremum and infimum (1 Viewer)

BenHowe · May 4, 2017

Hey,

I don't understand how to do prove this questions in general. Here's an example question

Thanks heaps in advance. Got my mid sem coming up next week and need to get this stuff

InteGrand · May 4, 2017

BenHowe said:
Hey,

I don't understand how to do prove this questions in general. Here's an example question
View attachment 33919

Thanks heaps in advance. Got my mid sem coming up next week and need to get this stuff

$\noindent The set $T$ is bounded above (note that $y \leq 1$ for all $y\in T$, for example). Also, $T$ is bounded below (note that $y \geq 0$ for all $y \in T$). Since $0 \in T$, as $0 = \frac{0^{2}}{0^{2} + 4}$ and $0 \in \mathbb{R}$, we have that $0$ is in fact the minimum of $T$, which implies it is the infimum of $T$. We also have that $1$ is the supremum of $T$. We already said $y \leq 1$ for all $y \in T$ (this follows from the fact that $x^{2} \leq x^{2} + 4$ for all $x \in \mathbb{R}$). Also, for any $\varepsilon > 0$, there is a $y\in T$ such that $y > 1- \varepsilon$ (you should be able to show this). This will allow us to conclude that $1$ is the supremum of $T$.$

$\noindent It might help your intuition if you can visualise or sketch the graph of $y = \frac{x^{2}}{x^{2} + 4}$.$

BenHowe · May 4, 2017

InteGrand said:
$\noindent The set $T$ is bounded above (note that $y \leq 1$ for all $y\in T$, for example). Also, $T$ is bounded below (note that $y \geq 0$ for all $y \in T$). Since $0 \in T$, as $0 = \frac{0^{2}}{0^{2} + 4}$ and $0 \in \mathbb{R}$, we have that $0$ is in fact the minimum of $T$, which implies it is the infimum of $T$. We also have that $1$ is the supremum of $T$. We already said $y \leq 1$ for all $y \in T$ (this follows from the fact that $x^{2} \leq x^{2} + 4$ for all $x \in \mathbb{R}$). Also, for any $\varepsilon > 0$, there is a $y\in T$ such that $y > 1- \varepsilon$ (you should be able to show this). This will allow us to conclude that $1$ is the supremum of $T$.$

$\noindent It might help your intuition if you can visualise or sketch the graph of $y = \frac{x^{2}}{x^{2} + 4}$.$

Hey, thanks for the reply.

I get that but I have to prove it with this argument. Ill screenshot it so we're on the same page. Also my calculus lecturer is a big fan of contradiction.

Thanks heaps for the help!

InteGrand · May 4, 2017

BenHowe said:
Hey, thanks for the reply.

I get that but I have to prove it with this argument. Ill screenshot it so we're on the same page. Also my calculus lecturer is a big fan of contradiction.

View attachment 33920

Thanks heaps for the help!

$\noindent Since $x^{2} \leq x^{2} + 4$ for all $x \in \mathbb{R}$, we have $\frac{x^{2}}{x^{2} + 4} \leq 1$ for all $x \in \mathbb{R}$, which implies $y \leq 1$ for all $y \in T$. Also, given any $0 < \varepsilon < 1$, take $x$ to be any real number with $x > \sqrt{\frac{4}{\varepsilon} - 4}$, e.g. take $x\equiv x(\varepsilon) := \sqrt{\frac{4}{\varepsilon} - 4} + 1$. Then $x \in \mathbb{R}$ since $\frac{4}{\varepsilon} - 4 > 0$ and $\frac{x^{2}}{x^{2} + 4} > 1 - \varepsilon$, as you should be able to show. This will show that $1$ is the supremum of $T$.$

Help with proving supremum and infimum (1 Viewer)

BenHowe

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InteGrand

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BenHowe

Active Member

InteGrand

Well-Known Member

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