show $ x \to (x,f(x))$ is an embedding Let $f:\mathbb{R}^n \to \mathbb{R}^m$ be ca continuous function and $(\mathbb{R}^n,\mathcal{O}_{\mathbb{R}^n}) $ the real numbers equipped with the euclidian topology.  Why the map $F:\mathbb{R}^n \to \mathbb{R}^{m+n}$ with $ x \to (x,f(x))$ is an embedding?
Of course $F$ is naturally injective and F is also continuous. But it's not clear to me why the map $F: (\mathbb{R}^n,\mathcal{O}_{\mathbb{R}^n}) \to (F(\mathbb{R}^n),\mathcal{O}_{\mathbb{R}^{m+n} | \mathbb{R}^n})$ is open? Where $\mathcal{O}_{\mathbb{R}^{m+n} | \mathbb{R}^n}$ is the subspace topology from $\mathbb{R}^n$ in $\mathbb{R}^{m+n}$.  Can someone give me a little hint?

It is easier to show that it is closed.
Indeed, let $A$ be closed in $\mathbb{R}^n$. We will show that $F(A)$ is closed. Pick any convergent sequence $(X_n)$ in $F(A)$. Obviously $X_n=(x_n, f(x_n))$ for some sequence $(x_n)$ in $A$. Since $X_n$ is convergent, then so is $(x_n)$ because projection to the first coordinate is continous. Since $A$ is closed then $(x_n)$ converges to $x\in A$ and since $f$ is continous then $f(x_n)$ converges to $f(x)\in f(A)$. Therefore $(x_n, f(x_n))$ converges to $(x, f(x))\in F(A)$ so $F(A)$ is closed.
Thus $F$ is both closed and bijective on image. Therefore it's an embedding.
