1. example, closed pointset

Give an example of a closed pointset $F \subseteq \mathcal{N}$ and a continuous $f: \mathcal{N} \rightarrow \mathcal{N}$, such that the image $f[F]$ is not closed.

I can not seem to come up with a suitable example. I need help on this problem. In our book, pointsets are subsets of Baire space. Also, $\mathcal{N}$ denotes Baire space. Thanks in advance.

2. Originally Posted by eskimo343
Give an example of a closed pointset $F \subseteq \mathcal{N}$ and a continuous $f: \mathcal{N} \rightarrow \mathcal{N}$, such that the image $f[F]$ is not closed.

I can not seem to come up with a suitable example. I need help on this problem. In our book, pointsets are subsets of Baire space. Also, $\mathcal{N}$ denotes Baire space. Thanks in advance.
$\mathbb{R}$ is a Baire space and it's not hard to do what you asked there. Does that help or do you need me to write out a specific example?

3. Could you write out a specific example (in $\mathbb{R}$ would be fine)? I still can't think of one there.

4. Originally Posted by eskimo343
Could you write out a specific example (in $\mathbb{R}$ would be fine)? I still can't think of one there.
Hmm, well here's maybe an easier one. $\mathbb{R}^2$ being a complete metric space is a Baire space, right? Now, Remembering $\mathbb{R}\approx\mathbb{R}\times\{0\}\subseteq\ma thbb{R}^2$use that and the fact that the canonical projection $\pi:\mathbb{R}^2\to\mathbb{R}x,y)\mapsto x" alt="\pi:\mathbb{R}^2\to\mathbb{R}x,y)\mapsto x" /> is continuous but $F=\left\{(x,y)\in\mathbb{R}^2:xy=1\right\}$ is closed but $\pi(F)=\mathbb{R}-\{0\}$. So, I think you can go from there.

5. Originally Posted by Drexel28
Hmm, well here's maybe an easier one. $\mathbb{R}^2$ being a complete metric space is a Baire space, right? Now, Remembering $\mathbb{R}\approx\mathbb{R}\times\{0\}\subseteq\ma thbb{R}^2$use that and the fact that the canonical projection $\pi:\mathbb{R}^2\to\mathbb{R}x,y)\mapsto x" alt="\pi:\mathbb{R}^2\to\mathbb{R}x,y)\mapsto x" /> is continuous but $F=\left\{(x,y)\in\mathbb{R}^2:xy=1\right\}$ is closed but $\pi(F)=\mathbb{R}-\{0\}$. So, I think you can go from there.
I think this is a common misunderstanding.
Drexel28 is speaking about Baire space as a topological property. (Roughly speaking, it is a topological space in which Baire category theorem holds.)
The original question of eskimo343 concerns the Baire space, which is a topological space on the set of all finite sequences of integers with the topology given in some sense by the tree structure. (It is homeomorphic to irrationals.)

6. Originally Posted by eskimo343
Give an example of a closed pointset $F \subseteq \mathcal{N}$ and a continuous $f: \mathcal{N} \rightarrow \mathcal{N}$, such that the image $f[F]$ is not closed.

I can not seem to come up with a suitable example. I need help on this problem. In our book, pointsets are subsets of Baire space. Also, $\mathcal{N}$ denotes Baire space. Thanks in advance.
I'll use the terminology I know from descriptive set theory, it should probably be almost the same as in Moschovakis' book Notes on set theory. (The OP is probably using this book, since the question appears as one of the problems in this book.)

Let L be the sets of all strings of length 1 with the exception of (0). I define $f: \mathcal{N} \rightarrow \mathcal{N}$ as follows:
$f(x)=(0)$, if $x\in L$ and
$f(x)=x$, otherwise.

Then $f(\mathcal N)$ is not a body of a tree, hence it is not closed. (Proposition 10.7)

The map f is continuous since:
$f^{-1} (\mathcal N_\emptyset) = \mathcal N \setminus L$ is open;
$f^{-1} (\mathcal N_{(0)}) = \mathcal N_{(0)} \setminus L$ is open;
$f^{-1} (\mathcal N_s) = \mathcal N_s \setminus L$ is open for $s\ne \emptyset,(0)$.
(I have used the definition of continuity using basic neighborhoods, perhaps another approach from Theorem 10.15 is also possible - I did not try that.)

7. Originally Posted by kompik
I think this is a common misunderstanding.
Drexel28 is speaking about Baire space as a topological property. (Roughly speaking, it is a topological space in which Baire category theorem holds.)
The original question of eskimo343 concerns the Baire space, which is a topological space on the set of all finite sequences of integers with the topology given in some sense by the tree structure. (It is homeomorphic to irrationals.)
Yes, I am sorry. I did not realize there was a "a" Baire space