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Is my logic correct or am I missing something?

Show that if $A\subseteq B$, then inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$

Case 1. If $A \subset B$ then there exists $b_1,b_2\in B$ such that $b_1,b_2 \notin A$. Let $a_1,a_2 \in A$ such that $a_1=$ inf A and $a_2$ = sup A. Suppose $b_1$=inf B and $b_2$=sup B than $b_1<a_1<a_2<b_2$ which implies inf $B<$ inf $A < $ sup $A<$ sup $B$

Case 2. If $A = B$ than every element in $A$ is in $B$. This implies that if $A$ is bounded above or below so is $B$ and vice versa. If the sup $B$ is defined to be the least upper bound and the inf $B$ is defined to be the greatest lowest bound. Than sup $B$ = sup $A$ and inf $B$ = inf $A$.

Since $A \subseteq B$ the following equality can be written as inf $B\leq$ inf $A\leq$ sup $A \leq$ sup $B$

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    $\begingroup$ You answered this question which asks the same question. $\endgroup$ – robjohn Oct 16 '13 at 15:31
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    $\begingroup$ On a purely pedantic note, you almost assume that $A$ is nonempty. Otherwise the statement doesn't hold. :) $\endgroup$ – user223391 Mar 30 '15 at 19:03
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Let's proceed by contradiction.

1) inf B $\le$ inf A

Assume, to the contrary, that inf B > inf A. Then there is some $x$ such that inf A < $x$ < inf B, but then $x \in A $ and $x \notin B$, a contradiction since $A \subseteq B$. (Also note here: If I were being super rigorous I would need to appeal to the definition of infimum to make sure that $x$ is indeed in $A$. Notice the strict inequality.) Thus, inf B $\le$ inf A

2) sup A $\le$ sup B Assume, to the contrary that sup A > sup B, then there exists some $x$ such that sup B < $x$ < sup A. However, then $x \in A$ and $x \notin B$. Again, a contradiction for the same reason as before.

3) inf A $\le$ sup A. By definition of sup and inf.

All together inf B $\le$ inf A $\le$ sup A $\le$ sup B.

Maybe someone can post a comment to help a little bit, what property of the real numbers am I using when I assume existence of $x$ like I do above? Certainly it must exist since there are no gaps in the real numbers. That way I can say that given two real numbers $a,b$ with $a <b$ then there exists $x$ such that $a <x <b$. Is this the Archimedian property? something topological?

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Your logic is incorrect in Case 1: You've assumed that $b_1 = \inf{B} \in B \setminus A$. Take $A = (0, 1)$ and $B = (0, 2)$ to see why this is quite problematic.


For a hint towards the correct direction, can you show that any lower bound of $B$ is a lower bound of $A$ as well? So the greatest lower bound of $B$ is still a lower bound for $A$?

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  • $\begingroup$ The sup and inf do not necessarily have to be in the set for example the sup $2-\dfrac{1}{n}$ is 2 since the limit of the sequence is 2. Is this correct? $\endgroup$ – adam Oct 16 '13 at 5:32
  • $\begingroup$ @adam Yes, a set does not have to contain its sup or inf. That is a counterexample, as are my $A$ and $B$. $\endgroup$ – user61527 Oct 16 '13 at 5:33
  • $\begingroup$ I rewrote case 1 to be this $\endgroup$ – adam Oct 16 '13 at 5:47
  • $\begingroup$ Case 1. If $A \subset B$ then there exists $b_1,b_2\in B$ such that $b_1,b_2 \notin A$. Let $a_1,a_2 \in A$ such that $a_1=$ inf A and $a_2$ = sup A. Let $b_1$ be a lower bound of $A$ and $b_2$ be a upper bound of $A$ Than there exist $b_3,b4$ such that for all $b \in B b_3<b$ and $b_4>b$. This implies that $b_3<b_1<a_1<a_2<b_2<b_4$ which implies inf $B<$ inf $A < $ sup $A<$ sup $B$ $\endgroup$ – adam Oct 16 '13 at 5:48
  • $\begingroup$ @adam $a_1$ and $a_2$ need not exist, as my example shows. There need not be any elements of $B$ which are lower bounds for $A$, so $b_1$ and $b_2$ might not exist either. $\endgroup$ – user61527 Oct 16 '13 at 5:49

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