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--- math104-s21:s:xingjiantao [2021/05/12 09:56]
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+++ math104-s21:s:xingjiantao [2026/02/21 14:41] (current)
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 ====== Xingjian Tao's Notes and Question List ======
 ===== Notes =====
-I made some notes on LaTex. You can see the PDF here {{ :math104:s:math104_notes.pdf |}}. The organization of definitions and theorems is mainly based on Rudin, though supplemented by Ross. Sometimes I tried to rewrite the contents in a more symbolic way, so there maybe minor mistakes.
+I made notes of some Chapters on LaTex. You can see the PDF here {{ math104-s21:s:math104_notes.pdf |}}. The organization of definitions and theorems is mainly based on Rudin, though supplemented by Ross. Sometimes I tried to rewrite the contents in a more symbolic way, so there maybe minor mistakes.
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 Answer: $\Longrightarrow$ can be found in 2.37 Theorem of Rudin. $\Longleftarrow$ can be found in Excercise 26 of Rudin Chapter 2.
+. If $f$ is a 1-1 continuous function on some interval $I$, prove $f^{-1}$ is continuous.
+Answer: 1-1 continuous on an interval $\Longrightarrow$ $f$ is strictly increasing or decreasing and continuous $\Longrightarrow$ $f^{-1}$ is continuous.
+. If $f$ is a 1-1 continuous function on an open interval $I$, prove that $f(I)$ is open.
+Answer:  According to Question 3, $f^{-1}$ is continuous. We know that $f^{-1}$ is continuous iff. for every open sets $O\subset I$, $f(O)$ is open. Since $I$ is open, we have $f(I)$ is open. (I think there maybe some mistakes so that I may only get $f(I)$ is open in $f(I)$ which is useless through my approach. But this proposition cannot be wrong since it's a statement in the proof of Ross 29.9 Theorem)\\
+Oh I See. We can first prove $f$ must be strictly montonic function and then $f(I)$ is open.

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