# subspace, closed map

• Jan 15th 2009, 12:06 PM
GenoaTopologist
subspace, closed map
Let $\displaystyle \pi_1:\mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)=x$, let $\displaystyle X$ be the subspace $\displaystyle (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}$

Prove $\displaystyle f=\pi_1|_X$ is a closed map but not an open map.
• Jan 15th 2009, 03:35 PM
Mathstud28
Quote:

Originally Posted by GenoaTopologist
Let $\displaystyle \pi_1:\mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)=x$, let $\displaystyle X$ be the subspace $\displaystyle (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}$

Prove $\displaystyle f=\pi_1|_X$ is a closed map but not an open map.

Hello, could you maybe clarify your notation? I am not exactly sure what this means, maybe some other more senior and knowledgable member recognizes this notation.
• Jan 15th 2009, 03:55 PM
chabmgph
Quote:

Originally Posted by GenoaTopologist
Let $\displaystyle \pi_1:\mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)=x$, let $\displaystyle X$ be the subspace $\displaystyle (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}$

Prove $\displaystyle f=\pi_1|_X$ is a closed map but not an open map.

Quote:

Originally Posted by Mathstud28
Hello, could you maybe clarify your notation? I am not exactly sure what this means, maybe some other more senior and knowledgable member recognizes this notation.

Though I am not used to this notation either, I am guessing it is to prove that
the function $\displaystyle f$ (which behaviors like $\displaystyle \pi_1$ but restricted in the subspace $\displaystyle X$) maps closed sets to closed sets but not open sets to open sets in $\displaystyle X$. (Thinking)
• Jan 15th 2009, 08:10 PM
Mathstud28
Quote:

Originally Posted by chabmgph
Though I am not used to this notation either, I am guessing it is to prove that
the function $\displaystyle f$ (which behaviors like $\displaystyle \pi_1$ but restricted in the subspace $\displaystyle X$) maps closed sets to closed sets but not open sets to open sets in $\displaystyle X$. (Thinking)

My main question is what does $\displaystyle (x,y)\mapsto \pi_1(x,y)=x$ mean? I have not seen that notation.
• Jan 15th 2009, 11:45 PM
Moo
Quote:

Originally Posted by Mathstud28
My main question is what does $\displaystyle (x,y)\mapsto \pi_1(x,y)=x$ mean? I have not seen that notation.

It just means that to (x,y), you give the value $\displaystyle \pi_1(x,y)$ (which is very logic since it's the image of (x,y) by the function pi_1 !)
And that you define $\displaystyle \pi_1(x,y)$ as being x.

It's a correct notation (Nod) And I'm sure you know what this function is (Rofl)
• Jan 16th 2009, 03:25 AM
Opalg
Quote:

Originally Posted by GenoaTopologist
Let $\displaystyle \pi_1:\mathbb{R} \times \mathbb{R}, (x, y) \mapsto \pi_1(x, y)=x$, let $\displaystyle X$ be the subspace $\displaystyle (\{0\}\times \mathbb{R})\cup (\mathbb{R}\times \{0\}) \subseteq \mathbb{R} \times \mathbb{R}$

Prove $\displaystyle f=\pi_1|_X$ is a closed map but not an open map.

The space X is just the union of the two coordinate axes. The mapping f (projection onto the first coordinate) acts as the identity on the x-axis, and takes everything on the y-axis to the origin. You should be able to see from that that the image under f of a closed set will always be closed (think of the set as the union of two parts, the part on the x-axis and the part on the y-axis: the image under f of each of these two parts is fairly obviously closed).

The map is not open, because for example the image of the open subset $\displaystyle \{0\}\times(0,1)$ of X is a single point and therefore not open.
• Jan 17th 2009, 01:54 PM
Mathstud28
Quote:

Originally Posted by Moo
It just means that to (x,y), you give the value $\displaystyle \pi_1(x,y)$ (which is very logic since it's the image of (x,y) by the function pi_1 !)
And that you define $\displaystyle \pi_1(x,y)$ as being x.

It's a correct notation (Nod) And I'm sure you know what this function is (Rofl)

Thank you Moo, I need to be careful using only one book since it has a different set of notation. (Hi) My book is not the only book.