Pedro Alcântara
@palcant.bsky.social
Moreno, alto, bonito, sensual e tesouro da mamãe.
Não acredito. O que mais tu detesta? Feriado? Paz entre os povos?
December 24, 2024 at 12:18 PM
Não acredito. O que mais tu detesta? Feriado? Paz entre os povos?
Eu me livrei de uma dor crônica no ombro com musculação também. Incrível.
November 19, 2024 at 6:54 PM
Eu me livrei de uma dor crônica no ombro com musculação também. Incrível.
Oh, I didn't catch that before! Well, this journal really has a controversial reputation. MDPI has been involved in polemics about its lack of serious review.
The paper is too long, so I would like someone to read and tell me what should be my opinion hahah
The paper is too long, so I would like someone to read and tell me what should be my opinion hahah
October 25, 2024 at 1:42 PM
Oh, I didn't catch that before! Well, this journal really has a controversial reputation. MDPI has been involved in polemics about its lack of serious review.
The paper is too long, so I would like someone to read and tell me what should be my opinion hahah
The paper is too long, so I would like someone to read and tell me what should be my opinion hahah
Note that the "global" statement is also false. Consider the continuous "step" function: f(x) = -1 for x\le 1; f(x) = x for -1\le x\le 1; f(x)=1 for 1\le x\le 2; f(x) = x-1 for 2\le x. Then f(-1/2, 3/2) = (-1/2, 1], which is not open in f(R) = [-1,\infty). But the restriction to V = (-1, 1) works.
October 18, 2024 at 2:34 AM
Note that the "global" statement is also false. Consider the continuous "step" function: f(x) = -1 for x\le 1; f(x) = x for -1\le x\le 1; f(x)=1 for 1\le x\le 2; f(x) = x-1 for 2\le x. Then f(-1/2, 3/2) = (-1/2, 1], which is not open in f(R) = [-1,\infty). But the restriction to V = (-1, 1) works.
Yes! And I'm working with germs. So, to be more precise, I want to prove the following statement: if f:U \to R^p is a continuous function defined on a open neighborhood of 0 in R^n with 0 as its only root, then there is an open neighborhood V of 0 contained in U such that f|_V: V \to f(V) is open.
October 18, 2024 at 2:22 AM
Yes! And I'm working with germs. So, to be more precise, I want to prove the following statement: if f:U \to R^p is a continuous function defined on a open neighborhood of 0 in R^n with 0 as its only root, then there is an open neighborhood V of 0 contained in U such that f|_V: V \to f(V) is open.
Sim, sim. E o pior foi que denunciei meio no automático, nem pensei em fazer o que a Letícia diz de mandar pro MP etc. Pelo menos agora fico mais ligado pra próxima.
September 24, 2024 at 4:21 PM
Sim, sim. E o pior foi que denunciei meio no automático, nem pensei em fazer o que a Letícia diz de mandar pro MP etc. Pelo menos agora fico mais ligado pra próxima.