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For context, I'm taking an introductory real analysis course, and our current topic is intervals. One of the questions requires to prove that, for every $x, y$ in some real interval $I$, with $x<y$, the interval $[x,y]$ is also contained in $I$.

I'm taking a shortcut by proving it's a subset of an open interval $I = (a,b)$, then arguing that $[x,y] \subset (I \cup G)$, for some arbitrary set $G \subset \mathbb R$. From there, changing $G$ should span all the cases I'm looking for.

My question: is the provided argument sufficient? Is it a non-trivial fact in set theory? Or am I wrongly using my thesis to prove the hypothesis? Thanks in advance!

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dayhhhdreaming is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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    $\begingroup$ I wouldn't say it's a shortcut, since the proof for the claim is pretty straightforward. Besides, at best $[x,y] \subseteq I \cup G$ only lets you claim the elements of $[x,y]$ are in one or the other. I suppose if you can prove it for all $G \subseteq \mathbb{R}$, somehow, then the case $G = \varnothing$ gets you the end result. But it seems roundabout at best and far from the intention of the exercise. $\endgroup$ – Eevee Trainer Feb 22 at 17:50
  • $\begingroup$ @EeveeTrainer Thank you for the comment! My goal with this is to prove that $[x,y]$ is further contained in $I = (a,b)$ but closed on one side, closed on both sides, and in $(a, +\infty)$ or $(-\infty, b)$. In my opinion, it would be less efficient to repeat the proof six times... $\endgroup$ – dayhhhdreaming Feb 22 at 17:59
  • $\begingroup$ My argument would just be to note that $$[x,y] = \{ z \mid z = (1-t)x + yt \text{ for some } t \in [0,1] \}$$ With this definition of an interval, one need only restrict $t$ a bit to show that an interval contains subintervals. $\endgroup$ – Eevee Trainer Feb 22 at 18:24
  • $\begingroup$ @EeveeTrainer This is a very unique way of defining this interval! I will consider it and try again :) $\endgroup$ – dayhhhdreaming Feb 22 at 19:44
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I'm going to answer the question in your title, and I hope that it gives you some insight. Yes, a subset is contained in any union of its superset with a different set.

I would say this is a trivial set theoretic fact, however if you don't trust a statement it is always worth going about a proof for it because maths is all about proving to ourselves that we're right. Here's a proof for it:

Lets take $A \subset B$ but $A \not \subset B \cup C$ for some set $C$. This implies there exists $a \in A$ such that $a \in B$ (by the first fact) but $a \not \in B \cup C$. This is a very clear contradiction.

I question you saying you're going to "prove it's a subset of the open interval $I = (a,b)$." Surely doing that is what you're trying to do? You can then just union on the end points to get the half open and closed intervals.

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Vicvic38 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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    $\begingroup$ Thank you, this was absolutely what I was going for. I didn't make it clear, now that I notice, that this far I have already proven that $[x,y]$ is a subset of $I$, thus a subset of $I \cup G$. The gimmick was defining $G$ as one of the endpoints, both, or another interval that extends one side to infinity, to not repeat the proof many times. But now I'm sure there's room to do so! $\endgroup$ – dayhhhdreaming Feb 22 at 19:37
  • $\begingroup$ @dayhhhdreaming Absolutely brilliant. I mean what I said when I said that its always a good idea to question statements and see if you can prove them, even if someone else has said that they're trivial. $\endgroup$ – Vicvic38 Feb 22 at 19:41

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