MHB Proving Subsets of Intervals in $\mathbb{R}$

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In the discussion on proving subsets of intervals in $\mathbb{R}$, two main points are highlighted: first, if \(x, y \in I\) with \(x \leq y\), then the closed interval \([x,y]\) is contained within \(I\); second, for an open interval \(I\) and any point \(x \in I\), there exists a \(\delta > 0\) such that the interval \([x-\delta, x+\delta]\) is also contained in \(I\). Participants are encouraged to share their progress on these proofs to facilitate better assistance. One user references a related exercise about finding a number between two given real numbers, indicating a connection to the topic. The discussion emphasizes the importance of understanding the properties of intervals in real analysis.
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Let $I \subseteq \mathbb{R}$ be an interval. Prove that

1. If $x, y \in I$ and $ x \le y$ then $[x,y] \subseteq I$.

2. If $I$ is an open interval, and if $x \in I$, then there is some $\delta > 0 $ such that $[x-\delta, x+\delta] \subseteq I$.
 
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Hello and welcome to MHB, NoName! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
 
Hello greg1313, and thanks for the warm welcome. So far I've figured that I'm probably supposed to use part (3) of http://mathhelpboards.com/analysis-50/using-axioms-ordered-field-18031.html#post82951 exercise.

If I understand correctly, that exercise says given any two real numbers $a, b$ such that $a < b$ we can always find a third $c$ such that $a < c < b$ and in particular $c = \frac{1}{2}(a+b)$, but I'm unable to prove it. I'll post an update if I make a progress/figure it out however.
 

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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