A question regarding quotient space

[yifli]

Does quotient space V/N include N itself?

I think it does because:

V/N includes all the cosets of N, and a coset of N is defined as N+$$\alpha$$={$$\epsilon$$+$$\alpha$$: $$\epsilon \in N$$}. This definition is from Advanced Calculus by Loomis & Sternberg and it does not say $$\alpha$$ cannot be 0. So V/N includes N itself when $$\alpha=0$$.

That said, I think V/N is equal to V.

Am I wrong?

Thanks

 

[micromass]

Does quotient space V/N include N itself?

Yes, but in the sense that $$N\in V/N$$, NOT in the sense that N is a subset of V/N!

That said, I think V/N is equal to V.

Am I wrong?

I'm afraid so, yes. The elements of V/N are subsets of V. So there's no way that V could ever equal V/N! (it could be isomorphic though, but you're not talking about that here).

For example, if $$V=N=\mathbb{R}$$, then $$V/N=\{\mathbb{R}+\alpha~\vert~\alpha\in \mathbb{R}\}=\{\mathbb{R}\}$$.
So V has infinite number of elements and V/N has only 1 element...

 

[wisvuze]

You have to keep in mind what the elements/vectors of V/N are. The vectors in this space are entire sets of their own. If V/N is your quotient space, then N is in the space, acting as the "zero vector"

 

[Bacle]

In addition, by , I think Lagrange's theorem, |V/N|=|V|/|N| , so the only way you
could have some kind of isomorphism between V and V/N is if |N|=1.

Maybe a good way of seeing things is that you have enlarged the identity
element in your group ( or other algebraic object) , from the identity element
to the whole subgroup (ideal, etc.) N.