# How to show that linear applications are closed and open [closed]

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

Is $f$ closed? Why?

Is it open? Why?

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

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## 1 Answer

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.

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