-1
$\begingroup$

Let $f: R^n\mapsto R^m$ a linear application.

Is $f$ closed? Why?

Is it open? Why?

$\endgroup$

closed as off-topic by José Carlos Santos, TheGeekGreek, jvdhooft, Saad, Shailesh Apr 9 '18 at 8:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, TheGeekGreek, jvdhooft, Saad, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

The projection $\pi_1: \mathbb{R}^2 \to \mathbb{R}$ is not closed.

Linear maps are open onto their image (open mapping theorem), between Euclidean spaces. Otherwise consider $f(x,y) = (x,0)$ on the plane.

$\endgroup$
  • $\begingroup$ The open mapping theorem requires surjectivity. For example $f: \mathbb R \to \mathbb R^2$ where $f(x) = (x, 0)$ maps $\mathbb R$ to the $x$-axis in $\mathbb R^2$ which is not open. $\endgroup$ – bitesizebo Apr 9 '18 at 7:34
  • $\begingroup$ Yes the linear function $f: \mathbb{R} \to \mathbb{R^2}$ with $f(x)=(x,0)$ is not open. $\endgroup$ – dem0nakos Apr 9 '18 at 7:36
  • 1
    $\begingroup$ @dem0nakos Thx,I added an example. In topology open often means open wrt the image space. $\endgroup$ – Henno Brandsma Apr 9 '18 at 7:37
  • $\begingroup$ @HennoBrandsma Oh , okay ! np :) $\endgroup$ – dem0nakos Apr 9 '18 at 7:39
  • $\begingroup$ Why is $\pi_1$ not closed? I am struggling to think of a counterexample. $\endgroup$ – Jsevillamol Jun 18 '18 at 8:47

Not the answer you're looking for? Browse other questions tagged or ask your own question.