Let $V$ be the set of sequences whose terms are contained in $\mathbb{R}^n . V$ is the set of functions $x(·) : N → \mathbb{R}^n $ which we denote as $\{x_n\}_n \subset \mathbb{R}^n$. $V$ is a vector space. $V_0$ is the subset of $V$ which consists of all convergent sequences, ie:

$V_0 := \{\{x_n \}_n ∈ V : \exists \space \space x ∈ \mathbb{R}^n | x = lim_{n→∞} x_n \}$

Let $A\subset \mathbb{R}^n$ and $V(A):= \{\{x_n\}_n \in V_0 : lim_{n→∞}x_n \in A\}$

Prove that if $A$ is affine, then $V(A)$ is affine?

for a set to be affine we need for the set to contain a line through any two distinct points $x_1 x_2$ in the set.

for a function to be affine, we need it to be of the form $f:\mathbb{R}^n \to \mathbb{R}^m, \space \space \space \space f(x) = Ax+b$

So is the only thing I need to do is show that $V(A)$ is an affine functionif it is written in the form $Ax + b$ if $A$ holds definition of an affine set?

I don't quite see how I would approach this, any help would be very appreciated, especially If I have misunderstood the question.


Suppose $A$ is affine, and let $\{x_n\}_n$, $\{y_n\}_n$ be two sequences in $V(A)$. Then $x_n \to x \in A$ and $y_n \to y \in A$, so since $A$ is affine, the set $\{ \lambda x + (1-\lambda) y \, | \, \lambda \in \mathbb R \}$ is contained in $A$, and in particular the sequence $\lambda x_n + (1-\lambda) y_n$ converges to $\lambda x + (1-\lambda)y$, so $\{ \{\lambda x_n + (1-\lambda)y_n \} \} \subseteq V(A)$, so $V(A)$ is affine.

I don't know why you mention affine functions.

Hope that helps,

  • $\begingroup$ Thanks, I should have read the question more carefully. For some reason I concluded $V(A)$ was a function, not a set. Thank-you for your help! $\endgroup$ – elbarto Jan 19 '15 at 19:30
  • $\begingroup$ @elbarto : Would have been probably very problematic to deal with this concept since $V(A)$ is not finite-dimensional! $\endgroup$ – Patrick Da Silva Jan 20 '15 at 1:56
  • $\begingroup$ Silly me! haha :P $\endgroup$ – elbarto Jan 20 '15 at 3:45

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