Question

If A and B are nonempty subsets of the Real Numbers, with A subset B, prove that

Inf B <= Inf A <= Sup A <= Sup B

Answer 1

Now I know that this is an obvious solution but I am not sure where to start.

Answer 2

Since \(A \subset B \iff \forall a: (a\in A \implies a \in B )\)
Then \( \inf A \in B \)
Either \( \inf A = \inf B\) or \( \inf A \neq \inf B\)
If \( \inf A < \inf B \implies \exists a \in A \) s.t. \( a \notin B\)
Which implies a contradiction and therefore \(\inf A \geq \inf B\)

The second inequality is by definition, and the third can be shown in similar fashion.

Answer 3

Since A and B are well ordered, we can look at A's infimum in terms of all the elements of B. Then by definition we get that \(\inf A \geq \inf B\)

Answer 4

inf A does not have to belong to the set B

Answer 5

A is a subset of B.

Answer 6

so...my point still stands