推 srewq :咦? 你有考慮小於8嗎? 不然 3.5^2 + 3.5 -12 = 3.75 02/11 02:41
→ srewq :3.5應該也要在A裡面吧~ 02/11 02:41
※ 編輯: GUOpp 來自: 61.229.231.105 (02/11 07:50)
→ GUOpp :謝謝s大的指正，已改過 02/11 07:51
推 rayar :謝謝兩位高手! 02/11 10:30
※ 引述《rayar (..........)》之銘言：
: Open subset
: 1. Let A ={ x ∈ R : 0 < x^2+x-12 < 8}.
: Show that A is an open subset of R.
x^2+x-12 = (x+4)(x-3)
x^2+x-20 = (x+5)(x-4)
所以只要證明 A = { x in R : -5 < x < -4 or 3 < x < 4 } 就可以了
: 2. Suppose U is an open subset of R and let B = { x ∈ R : y=2x ∈ U }
: Show that B is an open subset of R.
考慮此映射 f: B → U with f(x)=2x ，
證明 f 是 well-defined , bijection , continue，
再用“連續函數對開集的pre-image仍是開集”這個定理就可以證完了。
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