Proving (a,b) Open and [a,b] Closed

Real Analysis · Axiom Academy

EXAMPLE Proving (a,b) Open and [a,b] Closed Working straight from the definitions of open and closed sets in . Let a < b be real numbers. Prove that the open interval (a,b) is an open subset of , and that the closed interval [a,b] is a closed subset of . The picture behind both proofs Left: every interior point of (a,b) carries a little ball that stays inside, using radius . Right: every point outside [a,b] carries a little ball that misses [a,b] entirely, so the complement is open. You proved both halves straight from the definitions — one set is open because every point has room to spare, the other is closed because everything outside it has room to spare. Open means room at every point: a set is open when each of its points sits inside some ball that the set contains. The radius trick for (a,b) : take . Since a < x < b makes both gaps positive, r > 0 , and choosing the smaller gap keeps the ball from spilling past either endpoint. Closed = complement open: rather than chase limit points, show is open. A witness ball outside: a point p < a uses s=a-p ; a point q > b uses s=q-b . Either ball misses [a,b] entirely. Same idea, mirrored: "open" needed a ball inside the set; "closed" needed a ball inside the complement. The / gap-to-the-edge move powers both. These two arguments are the template for nearly every open/closed proof you will meet in a first real-analysis course.

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