Real Analysis – Intersection of Open Dense Subsets G_n of R^k is Not Empty and Dense

real-analysis

I am attempting to prove a special case of Baire theorem (ex. 30 in Rudin book).
It states: If $G_n$ is a dense open subset of $\mathbb{R}^k$ for $n = 1,2,3,\dots$, then $\bigcap_nG_n\neq \varnothing$ (in fact its dense in $\mathbb{R}^k$).

Here is my proof:

  1. Take some $V\subset \mathbb{R}^k,V$ is open. Form $U_1 = V\cap G_1$.
    It can be shown that $U_1\neq\varnothing$ and $U_1$ is open in $\mathbb{R}^k$
  2. Take some $p\in U_1$ then $\exists B_1\subset U_1,B_1$ is an open ball around $p$. It is possible to show $\exists C_1\subset B_1, C_1$ is closed ball around $p$. (I omit the poof to save the space) $C_1$ is compact since it is also bounded. Also $C_1\subset G_1$.
  3. Suppose $B_n$ has been constructed with $C_n\subset G_n$
  4. Then $U_{n+1}=B_n\cap G_{n+1}$
  5. By construction defined in (1) and (2) $\exists C_{n+1}\subset G_{n+1},C_{n+1}$ is compact
  6. By (5) $C_{n+1}$ satisfies our induction hypothesis
  7. The resulting set of nested compact sets is not empty by finite intersection property.
  8. Since $\forall n.C_n\subset G_n$ ,we can conclude that $\forall n.\bigcap_nG_n\neq\varnothing$

Q.E.D

Question:
How can I show that intersection not only non-empty it is also dense in $\mathbb{R}^k$? (Maybe my proof is not the best way)

I appreciate any help!
Leon

Best Answer

What you did already proves this! Notice that by construction the point you found lies on $V$. You can take $V$ to be an arbitrarily small neighborhood around any point you choose and repeat the procedure to find an element of $\cap G_n$ sufficiently close to that point. Note that it was crucial that the $G_n$ were dense in order to assert that the intersections were not empty.

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