MHB Another supremum and infimum problem

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For non-empty subsets S and T of the real numbers, if every element s in S is less than or equal to every element t in T, then the supremum of S is less than or equal to the infimum of T. The existence of both sup(S) and inf(T) is guaranteed because S and T are non-empty. Each element s in S serves as a lower bound for T, leading to the conclusion that s is less than or equal to inf(T). If sup(S) were greater than inf(T), it would imply the existence of an s in S that exceeds inf(T), resulting in a contradiction. Thus, the relationship sup(S) ≤ inf(T) holds true.
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Let $S$ and $T$ be non-empty subsets of $\mathbb{R}$, and suppose that for all $s\in S$ and $t\in T$, we have $s\le t$. Prove that $\sup S\le\inf T$.
 
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Alexmahone said:
Let $S$ and $T$ be non-empty subsets of $\mathbb{R}$, and suppose that for all $s\in S$ and $t\in T$, we have $s\le t$. Prove that $\sup S\le\inf T$.
Clearly $ \inf(T)$ & $\sup(S) $ exist. WHY?

$\forall s\in S$ we know $s\le\inf(T)~.$ WHY?$
 
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Plato said:
Clearly $ \inf(T)$ & $\sup(S) $ exist. WHY?

Because $S$ and $T$ are non-empty subsets of $\mathbb{R}$.

$\forall s\in S$ we know $s\le\inf(T)~.$ WHY?

I don't know. Why?
 
Alexmahone said:
I don't know. Why?
Well every $s\in S$ is a lower bound for $T$.
Therefore $s\le\inf(T)$.
 
Plato said:
Well every $s\in S$ is a lower bound for $T$.
Therefore $s\le\inf(T)$.

So if $\sup S>\inf T$, there must be an $s\in S$ such that $s>\inf T$. (Otherwise, $\inf T$ is a smaller upper bound for $S$ than $\sup S$.) So we get a contradiction. (Is that correct?)
 
Alexmahone said:
So if $\sup S>\inf T$, there must be an $s\in S$ such that $s>\inf T$. (Otherwise, $\inf T$ is a smaller upper bound for $S$ than $\sup S$.) So we get a contradiction. (Is that correct?)
Yes. It is correct.
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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