Question

(a) A function $f: X \rightarrow Y$ is called an open map if it maps every open set in $X$ to an open set in $Y$ and a closed map if it maps every closed set in $X$ to a closed set in $Y$. Provide an example of a function $f: \mathbf{R} \rightarrow \mathbf{R}$ which is continuous but not open and one which is continuous but not closed. (b) Prove that if a function $f: X \rightarrow Y$ is bijective and open, then it is also closed, and that if it is bijective and closed it must also be open.

    (a) A function $f: X \rightarrow Y$ is called an open map if it maps every open set in $X$ to an open set in $Y$ and a closed map if it maps every closed set in $X$ to a closed set in $Y$. Provide an example of a function $f: \mathbf{R} \rightarrow \mathbf{R}$ which is continuous but not open and one which is continuous but not closed.
(b) Prove that if a function $f: X \rightarrow Y$ is bijective and open, then it is also closed, and that if it is bijective and closed it must also be open.

Competitive Equilibrium: Theory and Applications

Bryan Ellickson 1st Edition

Chapter 4, Problem 14

Instant Answer

Step 1

This function is continuous because it is a polynomial function, but it is not open. To see this, consider the open set $(-1,1)$ in $\mathbf{R}$. The image of this set under $f$ is $[0,1)$, which is not open in $\mathbf{R}$. To provide an example of a function Show more…

Show all steps

Thanks for your feedback!

(a) A function $f: X \rightarrow Y$ is called an open map if it maps every open set in $X$ to an open set in $Y$ and a closed map if it maps every closed set in $X$ to a closed set in $Y$. Provide an example of a function $f: \mathbf{R} \rightarrow \mathbf{R}$ which is continuous but not open and one which is continuous but not closed. (b) Prove that if a function $f: X \rightarrow Y$ is bijective and open, then it is also closed, and that if it is bijective and closed it must also be open.

Play audio

Feedback

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later

Please give Ace some feedback