Honestly thinking about things in this way really helps
I can tell the difference. When I do proof problems from my linear algebra book, it normally doesn't take much time at all because everything seems much more straightforward. The thinking for these kinds of proofs just seems different for...
No worries! All of this is very helpful. I think I just need to stick it out for the simpler stuff, but also not be afraid to move along when necessary. The thing you said about definitions is something that has tripped me up more than once, no one has ever pointed it out to me before now.
Thanks for the advice! I don't know why I always feel the need to do every single part of a textbook. Just recently, someone told me that they normally only do half of the problems in a textbook. I had spent weeks doing every problem in every section of the books I'm working on and was wondering...
No this is like set theory stuff. In the book I'm using, the "real" stuff comes after set theory and properties of the real numbers. Most of the problems are just proofs of definitions similar to this. I've thought about skipping it multiple times but decided against it. I'm starting a degree in...
Ok, so here is what I have so far:
Suppose ##T_1## is infinite and ##\varphi : T_1 \rightarrow T_2## is a bijection.
Reasoning:
I'm thinking I would then show that there is a bijection, which would be a contradiction since an infinite set couldn't possibly have a one-to-one correspondence...
Let ##y \in E##. Assume that ##f## is surjective. There is some ##x \in f^{-1}(E)## such that ##y = f(x)##. So by definition, ##x =f(x) \in f(f^{-1}(E))##.
At least I finally got something understandable. I didn't have nearly as much trouble proving things about inverse images themselves (i.e...
Thank you for this. It took a while to reply because I had to get caught up on some school work.
Here's what I worked out:
If ##y \in f(f^{-1}(E))##, then ##y = f(x)## for some ##x \in f^{-1}(E)##. So if you have ##x \in f(f^{-1}(E))##, you have
it will clearly map back to the set ##E##. So we...
Thank you for this. It took a while to reply because I had to get caught up on some school work.
Here's what I worked out:
If ##y \in E##, then ##y = f(x)## for some ##x \in f^{-1}(E)##. Clearly, ##f^{-1}(E)## is the set of all points that map into ##E##, so ##f(f^{-1}(E))## will give us all...
ok, I've thought about it about. The goal of this proof should be to show that ##y \in f(f^{-1}(E) \iff y \in E##
So to start with ##y \in f(f^{-1}(E)##. ##y = f(x)## for some ##x \in f^{-1}(E)##
Or maybe since it is surjective, it is best to start with ##y \in E## so that we can show that...
Thank you for the reply! I always forget that to show equality after I have to clearly show that each set is a subset of the other :/
The only next step I can think of is that if f(x) is in f(E), then x is in E, right?
I typed this up in Overleaf using MathJax. I'm self-studying so I just want to make sure I'm understanding each concept. For clarification, the notation f^{-1}(x) is referring to the inverse image of the function. I think everything else is pretty straight-forward from how I've written it. Thank...
Here is my solution. I used mathjax to type it up in Overleaf. I feel like it makes sense, but I also have a feeling I might have "jumped the gun" with my logic. If it is correct, I would appreciate feedback on how to improve it. Thanks!