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1d
comment Boundedness in a topological space?
@mam I don't agree. We may disagree on the definition, but I would say Concept B is an extension of Concept A if there exists a universe where they are both defined and equivalent, or I would even say that, whenever they are both defined, they should then be equivalent. "Compactness" and "boundedness" are almost never equivalent (not even on the real line), and on metric spaces you can only prove that compact sets are bounded... I don't think it helps anyone to think that either is an extension of the other, it just adds to the confusion...
1d
comment Boundedness in a topological space?
@mma yes, open balls form a basis for the topology generated by the metric. But all your comments are aimed at metric spaces... there are no open balls in a general topological space... and no notion of boundedness
2d
comment Boundedness in a topological space?
@mma In a discrete metric space, all sets are closed and bounded, and a set is compact if and only if it is finite. Therefore, all infinite sets are bounded with non-compact closure.
Jun
28
comment Boundedness in a topological space?
@mma You're absolutely right, I can't believe this typo has been there all this time...
Jun
28
revised Boundedness in a topological space?
typo
Apr
28
awarded  Popular Question
Apr
17
revised Instructive proofs in functional analysis
clarified what looked like a confusing sentence
Mar
30
comment Constructive proof of the existence of an algebraic closure
@Rubertos You can show that there exists a set that contains all algebraic extensions up to isomorphism
Jan
11
comment What properties do you lose when you extend your number set?
@fvel If they don't fully answer your questions, you should explain why. Then maybe someone can answer whatever is left that you want to know.
Jan
9
revised What properties do you lose when you extend your number set?
edited tags
Jan
9
answered What properties do you lose when you extend your number set?
Dec
23
revised function such that $f(x\cdot t)=f(x)g(t)$
the title was confusing, and different from the statement of the question
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
@JaVaPG Yes, exactly. But as MPW (and Kaladin) mentioned, the two problems are basically equivalent (since $AB-I=-(I-BA)$).
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
@JaVaPG Correct.
Dec
23
revised Proof If $AB-I$ Invertible then $BA-I$ invertible.
improved formatting
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
@JaVaPG You're close, but that's not quite the answer. Recall that you are trying to prove $BA-I$ is invertible, not $BA$.
Dec
23
revised Proof If $AB-I$ Invertible then $BA-I$ invertible.
added 87 characters in body
Dec
23
comment Proof If $AB-I$ Invertible then $BA-I$ invertible.
I don't think this is a counter-example. I may have made a computation mistake, but I have $$AB-I = \begin{pmatrix}1&1\\-2&-2\end{pmatrix}$$ and also $$BA-I = \begin{pmatrix}0&1\\0&-1\end{pmatrix},$$ and neither are invertible. Moreover, you give a proof for when $A$ is invertible, which is the case in your "counter-example".
Dec
23
revised A counter example
it is more specific than general topology
Dec
23
revised Proof If $AB-I$ Invertible then $BA-I$ invertible.
the OPs question is not formulated as an "iff" statement