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Let A, B be non empty bounded sets on $\mathbb{R}$.

a) Show that $A \subseteq B \Rightarrow \inf(A)\ge \inf(B)$

b) Let $\lambda > 0$ show that $\inf(\lambda A)= \lambda \inf(A)$

c) Show that $\inf(A-B) = \inf(A) - \sup(B)$

d) Suppose that $\inf(A)=-2$ and $\inf(B)=3$, find $\inf(\{ 3a+2b \mid a\in A, b\in B\} )$

a) Since A and B are bounded sets, that means the infimum exists for both sets, let us define the infimum as $\inf(A)=a_o$ and $\inf(B)=b_o$. Also we assume $A \subseteq B$, so every element of $A$ is an element of $B$. Therefore by definition of an infimum $\inf(A)=a_0$ and $\inf(B)=b_0$, so since $A \subseteq B$ we can imply $\inf(A)\ge \inf(B)$.

b) We defined previously $inf(A)=a_0 \Rightarrow a_0 \le a \ \forall \ a\in A.$ Also by the distributive law, $\lambda A \Rightarrow \lambda a_0 \le a \le \lambda a_1 \ \forall \ a\in A \Rightarrow \inf(\lambda A) = \lambda a_0$. Therefore we can conclude $\inf(\lambda A) = \lambda \inf(A)$.

c) A-B is defined as the set of elements in A but not in B. I dont understand this. If I am understanding correct if $A =\{ a \in \mathbb{R}| 0 \le a \le 1 \}$ and $B =\{ b \in \mathbb{R}| 0.5 \le b \le 1 \}$ then $C = A-B =\{ c \in \mathbb{R}| 0 \le b < 0.5 \}$. Then $\inf(A) = 0$, $\sup(B) = 1$, and $\inf(A-B)= 0$ which does not satisfy the inequality.

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    $\begingroup$ I edited for proper use of $\inf$ rather than $inf$. It's coded as \inf. This has at least three effects: (1) It is not italicized, and (2) Proper spacing appears in expressions like $a\inf b$ and $a\inf(b)$ (note that there's more space after $\inf$ when no parentheses follow it, so spacing depends on surrounding text), and (3) This affects the posititions of subscripts, so that in a displayed (as opposed to inline) context you see this: $$ a \inf_{x\in S} b$$with the subscript $x\in S$ directly below $\inf.\qquad$ $\endgroup$ Commented Sep 8, 2016 at 23:22
  • $\begingroup$ C) if a in A and b in B then a-b >= inf A - sup B. So inf A - sup B is lower bound of A -B. For any e > 0 we can find inf A < a' < inf + e/2, a' in A and sup B - e/2 < b' < sup B, b' in B. So inf A - sup B < a'b - b' < inf A - sup B. So is greatest lower bound. $\endgroup$
    – fleablood
    Commented Sep 8, 2016 at 23:39
  • $\begingroup$ D) with b) you've prove inf 3A = -6 and inf 2b is 0. We can do a variation of c) to prove inf A + B = inf A + inf B. Indeed, you should have proven sup -B = - inf B and vice versa. $\endgroup$
    – fleablood
    Commented Sep 8, 2016 at 23:45

2 Answers 2

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The following will cost you marks on a test or assignment :

(a) You state correctly that there exist $a_0,a_1, b_0,b_1$ with $a_0\leq a\leq a_1$ and $b_0\leq b\leq b_1$ for all $a\in A$ and all $b\in B.$ However there are infinitely many such $a_0,a_1,b_0,b_1.$ E.g. if $A=(-1,2$) and $B=(0,1)$ we can have $a_0=b_0=-10$ and $a_1=b_1=12.$ ...... While it IS true when $a_0=\inf A$ and $b_0=\inf B,$ your statement that "by def'n of $\inf,$ we have $a_0\leq b_0$" is inadequate. It just says that what you have been asked to prove is true by def'n of $\inf,$ without explaining how.

(b). What you have shown is that if $a_0=\inf A$ then $\lambda a_0$ is a lower bound for $\lambda A.$ You have not shown it is the LARGEST lower bound for $\lambda A.$

For (a) let $x=\inf A.$ Then $ b\in B\implies b\in A\implies b\geq x.$ So $x$ is a lower bound for $B.$ So the largest lower bound for $B$ cannot be less than $x.$ That is , $\inf B\geq x=\inf A.$

Regarding (c) it appears that the Q defines $A-B=\{x-y: x\in A\land y\in B\}.$... I have seen & heard professional mathematicians say "$A$ minus $B$", meaning $A$ \ $B$, and write $A-B$ for $A$ \ $B$.

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Your proof could use some clarification.

Let's discuss (a). We know that bounded, nonempty sets have an infimum. You define $a_0, a_1, b_0,$ and $b_1$ as bounds for $A$ and $B$, and then state that, by definition, $a_1$ and $b_1$ are the greatest lower bounds. That is incorrect, because all you have said is that $a_1 \le a \forall a \in A$ (and hence is simply a lower bound, not necessarily the greatest lower bound. Moreover, you don't even need to go that path. Just appeal to the infimums of $A$ and $B$ directly.

Here's a proof for (a): Since $A$ and $B$ are nonempty, bounded sets, $inf(A)$ and $inf(B)$ exist. Then $inf(A) \le a \forall a \in A$, and since for every $b \in B, b \in A$, we have that $inf(A) \le b \forall b \in B$. Thus $inf(A)$ is a lower bound for $B$, so that $inf(B) \ge inf(A)$, since, by definition, $inf(B)$ is the greatest lower bound of $B$.

I'll let you try to patch up the other ones.

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