Do planes stop, or are they ever expanding? I am trying to understand sub-spaces in linear algebra and one of the rules mentions if W is my subspace then if k is any scalar and u is any vector in W then ku is in W.
I am unsure how this works ? Wouldn't there be a number I could use for the scalar that would make u so big it would  leave the subspace ?  i.e if my subspace was a plane in R3 wouldn't the vector off the edge of my plane? or are plane's edges ever expanding ? as in do the edges of a plane keep going in the direction they are going or do they stop ?
hope this makes sesne
 A: A plane in $\Bbb{R}^3$  has no edges, even though it seems like they do when we draw them. Your idea of "make it so big it would leave the subspace" is in direct opposition with the rule you just wrote. 
A: A plane in $R^3$ is infinitely large from the begining, it doesn't "expand" at all. There is no way then that any scalar multiplication can make it go outside of it
A: According to the definition of a subspace, what you have described is impossible. Namely, a subspace has to be close under scalar multiplication. In other words, if you take a vector $\bar v$ from a subset of a vector space, and you see that $k\bar v$ is not in that subset anymore, then by definition, that subset cannot possibly be a subspace.
A: To start off, you asked a fairly good question - namely because I haven't though about subspaces in the way you're asking. 
In any case, I'll take a shot at your question:
To make things simple, I'll assume the vector space we're working with to be in $R^3$ like you mentioned. Furthermore, I'll choose a plane - specifically the $X-Y$ plane - which is $\in R^2$. 
Since pictures make potentially abstract ideas easier to comprehend, I've included the following:

(I pulled it off Google somewhere).
As you can see, we have our vector space $R^3$, and our subspace - which we defined to be the $X-Y$ plane. Now things can get a little more abstract. A vector space, as you're obviously aware being that you asked the question, is defined to be a set that satisfies the following axioms (courtesy of Wolfram):

  
*
  
*Commutativity:
X+Y=Y+X.     
  
*Associativity of vector addition:
(X+Y)+Z=X+(Y+Z).     
  
*Additive identity: For all X,
0+X=X+0=X.   
  
*Existence of additive inverse: For any X, there exists a -X such that
X+(-X)=0.    
  
*Associativity of scalar multiplication:
r(sX)=(rs)X.     
  
*Distributivity of scalar sums:
(r+s)X=rX+sX.    
  
*Distributivity of vector sums:
r(X+Y)=rX+rY.    
  
*Scalar multiplication identity:
1X=X.

Now that you have a vector space defined, you can create vector spaces within vector spaces - also known as a subspace. Being that you already know that you have a valid vector space, the only two conditions that must be satisfied are closure under vector addition, and closure under scalar multiplication. Formally stated:

  
*
  
*W is nonempty, W≠∅.
  
*If x∈W and y∈W, then x+y∈W.
  
*If α∈C and x∈W, then αx∈W.
  

(Note that some people also formally include the condition that your subspace is non-empty). 
After all this, it is worth noting again that subspaces and vector spaces are very much definition driven. In other words, a set is only a subspace if and only if it satisfies the above axioms. 
Without looking at the image, we can see that if we have a subspace, then we know that any scalar we pick must be part of that subspace (i.e. it will never be "too" big as your wondering). Why? Because we have a subspace, which means that any vector that's an element of my subspace multiplied by a scalar $k$ will still be part of my subspace).
Considering the image, a basis for our vector space, $R3$, can simply be the standard basis: $\{(1,0,0)^T, (0,1,0)^T, (0,0,1)^T\}$. A basis for our subspace, then, would be $\{(1,0,0)^T, (0,1,0)^T\}$. As you can probably intuitively know, your X-axis can go on forever, your Y-axis can go on forever, and so forth. So your planes never end in this case (unless you define them to). 
Anyway, what's worth keeping in mind are the definitions of a vector space and subspace (especially once you start talking about function spaces). If you have a subspace, then you already know $ku$ will still be in the subspace.
Hope that helped!
A: You have not selected an answer yet, so here's a short explanation of why this is impossible strictly from a logic standpoint. We have the claim 

For all $k\in \mathbb{N}$ and $\mathbf{u} \in W$, we have $k\mathbf{u} \in W$

And you are also trying to introduce the claim 

There exists $k\in \mathbb{N}$ along with a $\mathbf{u}\in W$ such that $k\mathbf{u} \notin W$

These two claims contradict each other, in the same way the claim "$x$ is even and not even" is a contradiction. Since the first claim is part of the definition of a subspace, we go with it, and drop the second claim. Hence the second claim is necessarily false.
