0
$\begingroup$

I have the quotient space defined as

$$x+V = \{ y\in X \;:\; y-x\in V \}$$

Where $V\subset X$ are linear spaces. Since they are linear spaces they are closed under addition. So then shouldn't it follow that $x+V = V$ or $x+V = \emptyset$ depending on whether $x\in V$?

  • 0
    We have $x \in x+V$, since $x-x = 0 \in V$, so $x+V \neq \emptyset$. – 2017-02-12
  • 0
    @DustanLevenstein OK, so what I still don't understand is why $x+V\neq V$? – 2017-02-12
  • 0
    See what. You look locally. As $V$ is subspace of $X$. for every $x \in X\backslash V$, and $v \in V$, $v+x$ is defined. I mean in $X$. So you should treat it as a vector in $X$. – 2017-02-12
  • 0
    @kolobokish $x\in V$ and if $y\in V$ then $y-x\in V\implies y\in x+V$. If $y\notin V$ then $y-x\notin V\implies y\notin V$. So then $x+V = V$?? – 2017-02-12
  • 0
    For the next part, you are exactly right. What I mean, if $x\in V$, then if $y\in x+V$, then we can say there is some $v$, for which $y-x=v$, but being in $X$, we can say that $y=x+v$ then. And $x+v\ V$. So $x+V \subset V$, for the other part, as $V$ is linear space, for every element $v$ in $V$, $v-x\in V$, so $v\in x+V$, so $V \subset x+V$. – 2017-02-12
  • 0
    @kolobokish If $x+V = V$ then what's the point of defining this equivalence class? – 2017-02-12
  • 0
    It only the case, where $x \in V$. For the other cases, $x+V \neq V$. The idea od quotient is in aggregation into some bins to me, where we can treat the objects in bean as equivalent ones. And of course, when you take quotient by subspace $V$, then $V$ would be in some meaning a "core" of the beans, (in the "center" of that beans). And it can treated as 0 element. – 2017-02-12

1 Answers 1

0

First let me analyze a specific example: $X = \mathbb{R}^2 = \{(s,t) \, | \, s,t \in \mathbb{R}\}$ and $V = \{(r,r) | r \in \mathbb{R}\}$ which is just the line of slope 1 passing through the origin in $X$.

If $x=(0,-1)$ then for all $y=(s,t) \in X$ we have $y-x=(s-0,t+1)=(s,t+1)$ which is in $V$ if and only if $s=t+1$, and so we may take $t=s-1$ as the equation of the line $x+V$. This is just the line of slope $1$ with $t$-intercept $-1$.

Another general approach is to write $$x+V=\{x+v \,|\, v \in V\} $$ and in the special case above we get $$x+V = \{(0,-1) + (r,r) \,|\, r \in \mathbb{R}\} = \{(r,r-1) \,|\, r \in \mathbb{R}\} $$ which is a parameterization of the line $t=s-1$.

In fact, no matter what point you choose for $x$, the subset $x+V$ will be the line of slope $1$ passing through the point $x$.

Now, to what you asked in your question and the comment.

It is indeed true that if $x \in V$ then $x+V=V$. However, if $x \not\in V$ it is false that $x+V=\emptyset$, as my example shows specifically and as the comment of @DunstanLevenstein shows in general.

Also, in the comment you said "what I still don't understand is why $x+V \ne V$", and the answer is $x+V=V$ if and only if $x \in V$, an $x+V \ne V$ if and only if $x \not\in V$.

Finally, your title and your first sentence show some mis-understanding. It looks like you are confusing the quotient space with its elements. Your question does not give any particular notation for the quotient space itself, so I'm going to use one of the standard notations $X / V$.

So, it is improper to say that $x+V$ is "the quotient space".

What is proper is to say that "$x+V$ is an element of the quotient space $X / V$".

The key fact, which is hard to wrap one's head around at first, is that $x+V$ (which is a single object that happens to be a set) is an element of the quotient space $X/V$ (which is a set). In fact we can write $$X/V = \{x + V \,|\, x \in X\} $$

  • 0
    Thanks for the thorough explanation. What I was misunderstanding was: if $x\notin V$ and $y-x\in V$ then that would imply a contradiction because of the space being closed under addition. But $y\notin V$ as well so there is no contradiction. – 2017-02-12