T.1 · 実解析と位相幾何学
GRE Subject Test · GRE Mathematics トピック 5 2:14 英語ナレーション・英語＋中文字幕 burning-in
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Transcript
This unit applies the limit definitions for sequences and functions; distinguishes compactness, connectedness and completeness; and reasons in metric spaces and elementary topology.
本单元应用数列和函数的极限定义；区分紧致性、连通性和完备性；并在度量空间和初等拓扑中推理。
An epsilon-delta statement controls all sufficiently close inputs.
ε–δ 陈述控制所有足够接近的输入。
For x squared at 2: given any band within epsilon of 4, you must find a band within delta of 2 whose whole graph stays inside.
对 x² 在 2 处：任给 4 附近宽度为 ε 的带，你必须找到 2 附近宽度为 δ 的带，使这一段图像全部落在里面。
Epsilon comes first.
ε 先给定。
In real Euclidean space, closed and bounded means compact, but not in every metric space.
在实欧氏空间中，有界闭集就是紧集，但并非在所有度量空间中都如此。
Continuous images of compact sets are compact, so extrema are attained.
紧集的连续像是紧的，所以能取到最值。
Connectedness forbids a split into two open parts.
连通性不允许分成两个开的部分。
Completeness is a separate property.
完备性是另一回事。
f of x equals x on the open interval 0, 1 is continuous and bounded, but never equals its supremum 1.
f(x) = x 在开区间 (0, 1) 上连续且有界，但永远取不到上确界 1。
On the closed interval it attains its maximum at 1.
在闭区间上，它在 1 处取到最大值。
The missing endpoint is why the compact-domain theorem does not apply.
缺失的端点正是紧定义域定理不适用的原因。
A theorem's conclusion cannot be used before its hypotheses have been checked.
在检查定理的条件之前，不能使用它的结论。
Given epsilon, choose delta as the smaller of 1 and epsilon over 5.
给定 ε，取 δ = min(1, ε/5)。
Then x lies between 1 and 3, so x plus 2 is less than 5, and x squared minus 4 is less than 5 delta, at most epsilon.
于是 x 在 1 和 3 之间，所以 |x + 2| < 5，|x² − 4| < 5δ ≤ ε。
Separately, 1 to 3 is compact, so the minimum is attained: 1, at x equals 1.
另外， 是紧的，所以能取到最小值：在 x = 1 处为 1。
As a metric space by itself, 0, 1 is bounded and closed relative to itself.
作为自身的度量空间，(0, 1) 有界，且相对于自身是闭的。
But the Cauchy sequence 1 over n plus 2 has no limit in it, so it is not complete, hence not compact.
但柯西列 1/(n + 2) 在其中没有极限，所以它不完备，因此也不紧。
Heine-Borel cannot help: it needs closed as a subset of R.
海涅–博雷尔定理帮不上忙：它要求在 R 中是闭集。
Check yourself.
自我检测。
Epsilon comes first.
ε 先给定。
Closed and bounded gives compact in R n, not in every metric space.
有界闭集在 Rⁿ 中是紧的，在一般度量空间中则不一定。
x on the open interval never attains 1.
x 在开区间上取不到 1。
The open interval is not complete.
开区间不完备。
And a continuous image of a compact set is compact.
紧集的连续像是紧的。