to prove that Inf$(B) \sqsubseteq $ Inf$(A)$.

Question

In a poset $(P, \sqsubseteq)$, and $A \subseteq B \subseteq P$. Also Infimum$(A)$ and Infimum$(B)$ exists.

I need to prove that Inf$(B) \sqsubseteq $ Inf$(A)$.

My Attempt.

Considered $x \in A$, it means it is also there in $B$

So Inf$(A) \sqsubseteq x $, similarly Inf$(B) \sqsubseteq x $

But I am not able to conclude Inf$(B) \sqsubseteq $ Inf$(A)$

Does anybody have a hint.

[Edit]- My understanding is Inf$(A)$ is the lower bound for all elements of $A$

And similarly Inf$(B)$ is the lower bound for all elements of $B$

So from this I can conclude Inf$(B) \sqsubseteq $ Inf$(A)$ but How do I know that Inf$(A) \in A$, It can be outside of the A as well.

Answer 1

Hint: Show that $\operatorname{inf}(B)$ is a lower bound for $A$, and conclude your result from the definition of $\operatorname{inf}(A)$.