# Math Help - subspace, closed map

1. ## subspace, closed map

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

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

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

Prove $f=\pi_1|_X$ 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.

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

Prove $f=\pi_1|_X$ is a closed map but not an open map.
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 $f$ (which behaviors like $\pi_1$ but restricted in the subspace $X$) maps closed sets to closed sets but not open sets to open sets in $X$.

4. Originally Posted by chabmgph
Though I am not used to this notation either, I am guessing it is to prove that
the function $f$ (which behaviors like $\pi_1$ but restricted in the subspace $X$) maps closed sets to closed sets but not open sets to open sets in $X$.
My main question is what does $(x,y)\mapsto \pi_1(x,y)=x$ mean? I have not seen that notation.

5. Originally Posted by Mathstud28
My main question is what does $(x,y)\mapsto \pi_1(x,y)=x$ mean? I have not seen that notation.
It just means that to (x,y), you give the value $\pi_1(x,y)$ (which is very logic since it's the image of (x,y) by the function pi_1 !)
And that you define $\pi_1(x,y)$ as being x.

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

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

Prove $f=\pi_1|_X$ 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 $\{0\}\times(0,1)$ of X is a single point and therefore not open.

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

It's a correct notation And I'm sure you know what this function is
Thank you Moo, I need to be careful using only one book since it has a different set of notation. My book is not the only book.