# Is a subset contained in a union of its superset?

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 $$x, the interval $$[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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• 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. – Eevee Trainer Feb 22 at 17:50
• @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... – dayhhhdreaming Feb 22 at 17:59
• 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. – Eevee Trainer Feb 22 at 18:24
• @EeveeTrainer This is a very unique way of defining this interval! I will consider it and try again :) – dayhhhdreaming Feb 22 at 19:44

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.
• 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! – dayhhhdreaming Feb 22 at 19:37