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Let $R= [a,b] \times [c,d]$ be an arbitrary rectangle.

Define $S_1 =\{f(x,y): (x,y)\in R\} $ and $S_2=\{f(x,y):a\leq x \leq b\}$

Claim: For every $c\leq y \leq d$ we want to show that $\inf(S_1) \leq \inf(S_2)$.

My attempt: Let $y_0 \in [c,d]$ be arbitrary.

I realized that by drawing a picture the set $S_2$ the infimum of it is the fixed $y_0 $ we let from the beginning. However, I am not sure how to formally prove this whether if its by definition or by some theorem. Please help, thank in advance.

Clayton
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javacoder
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  • Please help!!!! – javacoder Feb 11 '19 at 05:23
  • Thank you so much, but if you don't mind could you confirm if this is correct set represeting $[a,b] \times {y}={(x,y): a \leq x \leq b,y = y}$ I'm just not 100% sure whether if $y = y$ makes sense... Please and thanks! – javacoder Feb 11 '19 at 05:44
  • What is the connection between $S_2$ and $y$s? Is $S_2 = S_2(y)$? Cause now it seems the claim should be exactly the opposite. – dEmigOd Feb 11 '19 at 07:21
  • You need to consider how you've defined $S_2$. As $S_1$ and $S_2$ are currently written, you have $S_1\subseteq S_2$ (to see this more clearly, consider drawing a picture of the domain of $f$ as it pertains to $S_1$ and $S_2$). If you mean for the domain of $f$ as it pertains to $S_2$ to be a fixed horizontal line, then you should write something like $S_2(y_0)={f(x,y):a\leq x\leq b,\ y=y_0}$. – Clayton Feb 11 '19 at 15:55

1 Answers1

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(1) You have the inequality backwards. For example if the domain of $f$ is $\Bbb R^2$ and $f(x,y)=y^2$ and $R=[0,2]\times [1,2]$ then $\inf S_1=\inf [1,4]=1>0=\inf [0,\infty)=\inf S_2=0.$

(2) $S_1\subset S_2$ so any lower bound for $S_2$ is also a lower bound for $S_1.$ That is, if $A$ is any lower bound for $S_2,$ then for any $z\in S_1$ we have $z\in S_2,$ so $z\ge A.$

So let $L$ be the greatest lower bound of $S_2$.Then $L$ is $one$ $of$ the lower bounds for $S_1,$ so the $greatest$ of the lower bounds for $S_1$ must be be at least $L$.

Clayton
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DanielWainfleet
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    @Clayton. Thanks. I have changed it. – DanielWainfleet Feb 11 '19 at 17:37
  • Sorry for the late response so is $[a,b] \times {y} \subset [a,b] \times [c,d]$ wrong? This hint (from @Clayton) actually helped me solve the problem that I needed it to be solved. Maybe the claim I wrote was unclear is that the case? – javacoder Feb 12 '19 at 03:20
  • The claim I made above I believe is correct, it should be $\inf(S_1) \leq \inf(S_2)$. I believe the inequality is not backwards I drew a picture to show that the claim makes sense. When we have for any $y\in[c,d]$ fixed the infimum of $S_2$ would just be at $y$. And infimum of $S_1$ would be below the above $y$.When we pick $y$ big enough (but smaller than $d$) then clearly for this $y$ the infimum of $S_1$ would be smaller. Say the infimum of $S_1$ is $c$ then we can always pick a $y > c$ and knowing that $y$ is the infimum $S_2$ we know $y>c$ from how we chose $y$ then our claim follows. – javacoder Feb 12 '19 at 03:26
  • @javacoder: your claim is correct if you assume $y\in[c,d]$ is fixed. The way you have phrased it in the original post is not super clear about this, though. As it is defined, $S_2$ does not have a restriction on $y$ (meaning that $y\in(-\infty,\infty)$ as opposed to $[c,d]$). – Clayton Feb 12 '19 at 05:06
  • I thought it was a mathematical/philosophical/logical convention that if quantifier of a "for all" statement was outside of a set it would imply that $y$ is "fixed" (or arbitrary as on paper it would look better). At least that's how I been taught. I would very appreciate if this topic can continue to be discussed will appreciate a follow up! @Clayton – javacoder Feb 12 '19 at 17:23
  • Sure, @javacoder. The specific thing here is that you need $y\in[c,d]$. If you look at your definition of $S_2$, you make no such assumption. (Note: you make this statement in your claim, but we must look to the definition of $S_2$ to make sure we are working with the right thing). – Clayton Feb 12 '19 at 17:51