How can I show $R^n$ is dense in $S^n$?

Question

How can I show $R^n$ is dense in $S^n$? I wanted to show $S^n$ is compactification of $R^n$. for this I need $R^n$ is not compact, for this there is no problem, and $S^n$ is compact, I did it with stereography, because it's covered by finite open set, Now I want to show $R^n$ is dense in $S^n$, for this maybe easier work with $(0,1)$ instead of $R$ because they are isomorphic, Is that correct? could you help me for further?

Answer 1

If $\Bbb R^n$ is realized as the obvious homeomorphic copy $X \subset S^n$ of it inside the one-point compactification $S^n$, then just note that any neighborhood of the point $\infty$ at infinity, in the one point compactification $S^n$, intersects $X$.

So $\infty$ is a limit point of $X\subset S^n$, and tautologically the only limit point as $S^n - \Bbb R^n$ contains nothing except $\infty$. Thus taking closure adds the point at infinity to $X$, hence $\text{cl}(X) = S^n$. That proves that $X$ is dense in $S^n$.

Answer 2

If $X$ is a topological space and $a\in X$ is not an isolated point, then $X\setminus\{a\}$ is dense in $X$. Indeed, any open subset $U$ of $X$ that fails to intersect $X\setminus\{a\}$ must ba a subset of $\{a\}$. As $\{a\}$ is not open, such $U$ must be empty. In other words, every non-empty open $U$ intersects $X\setminus\{a\}$.