Question

I need some help about topology homework.. anyone..

Answer 1

@helder_edwin

Answer 2

first one i can do
negate the definition by picking \(\epsilon =\frac{1}{2}\) then show that there does not exist a \(\delta\) such that \(|x|<\delta\implies \sin(\frac{1}{x})<\frac{1}{2}\)

Answer 3

that is because no matter how small \(\delta\) is there will be some \(k\) such that
\[\frac{2}{(2k+1)\pi}<\delta\] and for that \(k\) and all other larger you will have \(\sin(\frac{1}{x})=1\)

Answer 4

yes I believe we can choose epsilon, since it is negation..

Answer 5

thanks it makes sense to me..

Answer 6

sin(1/x) must be greater than epsilon, and it is true, how about 1/(2kpi)<delta
we have sin(1/x)=0 not greater than epsilon. can we use this..

Answer 7

@satellite73
hi, satellite73, can you check my comment here

Answer 8

hi
do you know exactly how to negate the definition of
\[\lim_{x\to a}f(x)=L ?\]that is what you need to use

Answer 9

definition of
\[\lim_{x\to a}f(x)=L \]
\(\forall \epsilon> 0\), \(\exists \delta \) such that \(|x-a|\implies |f(x)-L|<\epsilon\)

Answer 10

yes..

Answer 11

it should be I guess \[|x-a|\implies |f(x)-L|\ge \epsilon\]

Answer 12

negation is
\(\exists \epsilon >0\) such that \(\forall \delta\), \(\exists x\) such that \(|x-a|<\delta\) and \(|f(x)-L|>\epsilon\)

Answer 13

yes this is what I know..

Answer 14

you need to use what i wrote above (took me a while)
so the logic is this
you exhibit a specific \(\epsilon\) and then show that no matter how small \(\delta\) is, then you can find some \(x\) for which \(|x-a|<\delta\) and at the same time for that \(x\) you have \(|f(x)-L|>\epsilon\)
my suggestion was to pick \(\epsilon =\frac{1}{2}\) whatever you pick, it should be specific not general

Answer 15

in your example, you have to show that \(\exists \epsilon>0\) so that \(\forall \delta\) \(\exists x\) with \(|x|<\delta\) and \(|\sin(\frac{1}{x})|>\epsilon\)

Answer 16

yes ..

Answer 17

but it is gonna be contradiction right, and we are gonna say f(x) is not cont. at x=0..

Answer 18

since \(\sin(\frac{1}{x})=1\) infinitely often close to zero, you can show this
pick \(\epsilon=\frac{1}{2}\) for example, then \(\forall \delta>0\) \(\exists \frac{1}{(2k+1)\pi}<\delta\) and so \(\sin(\frac{(2k+1)\pi}{2})=1>\frac{1}{2}\)

Answer 19

you are negating the definition, but this is not a proof by contradiction, it is a direct proof, using the negation of the definition

Answer 20

a proof by contradiction would be to assume it is continuous and then arrive at some contradiction, but you do not need to do this here. write down exactly what the negation of the definition is, then prove it directly

Answer 21

can I use my example to get contradiction..sin(1/x)=0 but greater than 1/2

Answer 22

not greater than.

Answer 23

i should have said \(\forall \delta>0\)
\(\exists k\) such that \(\frac{1}{(2k+1)\pi}<\delta\)

Answer 24

it makes more sense now..

Answer 25

be careful here, you cannot write
\[\sin(\frac{1}{x})=0\] that makes no sense

Answer 26

again, i stress work the directly from the definition of " \(f\) is not continuous at \(a\)"
i know it seems like a proof by contradiction, but it is not
the definition of "\(f\) is not continuous at \(a\)" i wrote above

Answer 27

thanks.. it must be grater than 0

Answer 28

question 2 looks annoying but i think you can do it by saying something like this
for any two points you can join them by a straight line (you can even write down what the line would be) unless the line would contain those two points, in which case you can join them by a horizontal and vertical line that skips those points

probably an easier way, not sure

Answer 29

yeah, I thought something like that.. how about 5)ii)

Answer 30

for 5) ii) can we say that since A is compact it is finite.. I know it is bounded and closed.. is it true that every closed and bounded set is finite..