Proof about linear space, product equal to zero

[trenekas]

Hello! I have a problem with one proof. The task is:

Suppose that X is linear space, x belongs to X and λ is real number. Proof if λx=0 so λ=0 or x=0. And there are conditions. Can use only this properties μ also is real number:

I tryed to prove that but completely fails.
Let λ=0. According to d, λx=0
Let x=0. From c and g => λ(0+0)=λ0+λ0. And then i don't know what to do next. And also i think need to prove that if λ and x is not equal to zero when the product of them also not equal to zero.

Thanks for helping!

 

[jbunniii]

Hint: if $\lambda \neq 0$, then $\lambda$ has a multiplicative inverse.

 

[trenekas]

Hint: if $\lambda \neq 0$, then $\lambda$ has a multiplicative inverse.

Understand :) Thank you a lot

 

[trenekas]

sorry but after few hours i have little doubt about my proof so if i missunderstand just say.

if λ=0 then λx=0 according to (d).
if λ isn't equal to zero, then there is a multiplicative inverse. λx=0, λ not equal to zero and μ=1/λ . when μ(λx)=(μλ)x=1x=x. And there is one way, x=0. So if λx=0 then λ=0 or x=0.

its good or not? thanks for answer :)

 

[jbunniii]

sorry but after few hours i have little doubt about my proof so if i missunderstand just say.

if λ=0 then λx=0 according to (d).
if λ isn't equal to zero, then there is a multiplicative inverse. λx=0, λ not equal to zero and μ=1/λ . when μ(λx)=(μλ)x=1x=x. And there is one way, x=0. So if λx=0 then λ=0 or x=0.

its good or not? thanks for answer :)

Yes, I think the logic is right. You have shown that if $\lambda x = 0$ and $\lambda \neq 0$, then $x = 0$. But you should try to state it more clearly, and specify which vector space properties you are using. Something like the following:

Suppose $\lambda x = 0$ and $\lambda \neq 0$. Then $\lambda$ has a multiplicative inverse, call it $\mu = \lambda^{-1}$. By property (e), $\mu(\lambda x) = (\mu \lambda) x = 1x = x$, where the last equality is true because of property (d). But $\lambda x = 0$, so this means...

 

[trenekas]

Yes, I think the logic is right. You have shown that if $\lambda x = 0$ and $\lambda \neq 0$, then $x = 0$. But you should try to state it more clearly, and specify which vector space properties you are using. Something like the following:

Suppose $\lambda x = 0$ and $\lambda \neq 0$. Then $\lambda$ has a multiplicative inverse, call it $\mu = \lambda^{-1}$. By property (e), $\mu(\lambda x) = (\mu \lambda) x = 1x = x$, where the last equality is true because of property (d). But $\lambda x = 0$, so this means...

thanks dude very much. Its difficult to me all that write in english because my english isn't very good. in my native language i will try do the best. )

 

[trenekas]

Hello. I don't want to create another thread so i ask here.

There is task:

"A" is set of Euclidean space and x is "A" point of contact (x belong to Ā, don't know how is calling the set of all points of contact in english.). Need to prove that there is sequence of A elements which converges to x.

I don't have the idea how that prove. I found one theorem but it say that there is sequence of set elements which converges to x when x is borderline point.

If something not understand i'll try to say more cleary if it will be possible. Thanks.

 

[jbunniii]

Hello. I don't want to create another thread so i ask here.

There is task:

"A" is set of Euclidean space and x is "A" point of contact (x belong to Ā, don't know how is calling the set of all points of contact in english.). Need to prove that there is sequence of A elements which converges to x.

I don't have the idea how that prove. I found one theorem but it say that there is sequence of set elements which converges to x when x is borderline point.

If something not understand i'll try to say more cleary if it will be possible. Thanks.

It's better to create a new thread if you have a new question, unless it is very closely related to the existing thread.

In English, $\bar{A}$ is usually called the closure of $A$, and the elements of $\bar{A}$ are sometimes called the points of closure of $A$. The closure $\bar{A}$ consists of $A$ and all limit points of $A$.

Here is a hint to get you started: if $x \in \bar{A}$ then either $x \in A$ or $x$ is a limit point of $A$.

If $x \in A$ then it's trivial to create a sequence in $A$ which converges to $x$.

If $x$ is a limit point of $A$, then every neighborhood of $x$ contains an element of $A$ that is different from $x$. Use that fact to construct a sequence converging to $x$.

 

[trenekas]

omg :D thank you very much ) you are awesome!