How to show that a map in a metric space is open or closed?

Question

Show that the projection map $f_j$:$R^{n}$$\rightarrow$ R given by $f_j$(x) =$x_j$ is open. It is not closed. Honestly I don't have any work to show because i just know the defination of open map and closed map. The book i am solving they didn't provide any examples. I want help to understand this problem and there are many problems so i need to understand what i have to do when i am facing this type of problem.

Answer 1

An open map between topological spaces is a map that maps open sets to open sets. One can verify that a map $f:X\to Y$ between topological spaces is open if and only if it maps basic open sets to open sets.

In $\mathbb{R}^n$, we have a basis given by open balls $B_{\epsilon}(x)$, where $x$ is some point in $\mathbb{R}^n$. Specifically, this ball can be described as the following set: $$B_{\epsilon}(x)=\{y\in \mathbb{R}^n\mid d(x,y)<\epsilon\}$$In our case, the map $d(x,y)$ is just the Euclidean distance.

We will now show that this open ball is mapped to an open set in $\mathbb{R}$. We will assume without loss of generality that $x$ is the origin; to make computations easier. Note that in this case $y$ is in $B_{\epsilon}(x)$ iff $||y||<\epsilon$.

I claim that the following holds: $$f_j(B_{\epsilon}(x))=B_{\epsilon}(x_j)$$ Let $y$ in $B_{\epsilon}(x)$. Then $\sqrt{y_1^2+\ldots+y_j^2+\ldots+y_n^2}<\epsilon$ In particular, this states that $\sqrt{y_j^2}=|y_j|<\epsilon$. This shows that $f_j(y)$ is in $B_{\epsilon}(x_j)$.

Now for the other side, suppose that $z$ is in $B_{\epsilon}(x_j)$. That is, $|z|<\epsilon$. We can now consider the element $\tilde{z}:=(0,\ldots,z,\ldots, 0)$ in $\mathbb{R}^n$, where $z$ is on spot $j$, and see that this point $\tilde{z}$ is indeed in $B_{\epsilon}(x)$, and that $f_j(\tilde{z})=z$.

Hence, we see that the image of the open ball $B_{\epsilon}(x)$ is an open ball in $\mathbb{R}$, showing that the map is open.

Answer 2

To show that it is open, choose an open subset of $\mathbb{R}^n$ and show that its image under $f_j$ is still open. The Euclidean topology on $\mathbb{R}^n$ is equivalent to the product topology, hence we may consider an open subset of $\mathbb{R}^n$ to have the form $$\displaystyle\bigcup_{i_1, \ldots, i_n}U_{i_1}\times\ldots\times U_{i_n}$$ for open subsets $U_{i_1},\ldots, U_{i_n}$ of $\mathbb{R}$. Then observe that \begin{align*} f_j\left(\bigcup_{i_1, \ldots, i_n}U_{i_1}\times\ldots\times U_{i_n}\right)&=\bigcup f_j(V_{i_1}\times\ldots\times V_{i_n})\nonumber\\ &=\bigcup V_{i_j}\nonumber \end{align*}

where the last expression is an open subset of $\mathbb{R}$, being the union of open subsets.

To show it is not closed, consider for example the case $n=2$ and the subset of $\mathbb{R}^2$ given by $$F=\{(x,y)\in\mathbb{R}^2: xy=1\}$$ Then, calling $f_1$ the projection onto the first factor of $\mathbb{R}^2$, we get that $$f_1(F)=\mathbb{R}\setminus \{0\}$$ which is not a close subset of $\mathbb{R}$.