Prove the metric induced topology is the same as the product topology Let $P = \prod_{n=0}^\infty I_n , \ I_n = \left[0,1/n\right] $ and $d(\bar x,\bar y) = \left [ \sum_{n=0}^\infty (x_n-y_n)^2 \right]^{1/2} for\ \bar x,\bar y \in P.$ Prove that the topology induced by the metric d is exactly the same as the product topology. Deduce $I^\mathbb{N}$ is compact.
I know I have to proof that each topology is contained in the other but don't know how to proceed. Any help or hint would be appreciated. I'm also confused 
about the last part: how can I deduce from this equivalence that $[0,1]^\mathbb{N}$ is compact?
 A: Here are some hints.  You want to show that open sets in the product topology are open in the metric topology and vice versa.  The direct approach is to let $U$ be open in the product topology and let $x \in U$.  Then you need to find an $r>0$ such that $B(x,r) \subset U$, where $B(x,r)$ is the metric ball centered at $x$ with radius $r$.
General simplification: you may assume $U$ is a basic open set.
Simplification for the product topology: you can assume $U$ is an element of the subbasis consisting of basic open sets of the form $S_k = U_k \times \prod_{n \neq k} [0, 1/n]$, where $U_k \subset [0,1/k]$ is open.  The reason why we can do this is that a basic open set is (by definition) an intersection of finitely many subbasic sets.  Since there are only finitely many, you can choose $r > 0$ to be the minimum among the $r_k > 0$ satisfying $B(x,r_k) \subset S_k$.
So, your exercise here reduces to finding for each $x \in S_k$ an $r_k > 0$ such that $B(x,r_k) \subset S_k$.  You need to choose $r_k$ and this has to be related to $U_k$.  Also, you'll need an easy inequality.
For the opposite implication, the direct approach is to let $B(x,r)$ be a metric ball and let $y \in B(x,r)$ and to find a basic open set $U$ in the product topology such that $y \in U \subset B(x,r)$.
Simplification for the metric topology: it suffices to figure out how to do this when $x=y$.  Why?  In the metric topology, you can always choose $r' > 0$ such that $B(y,r') \subset B(x,r)$.  So, if you know how to find $U$ such that $U \subset B(y,r')$, you're done.
So, let $x \in P$ and $r > 0$.  And think about what happens for various (small) values of $r$.  You should be able to convince yourself that if $y \in P$ is such that $y_k$ is close to $x_k$ for $k = 0, \dots, N$, then $d(x,y) < r$ no matter the values of $y_k$ (because $|x_k-y_k| \leq 1/k$).  Figuring out what $N$ is in terms of $r$ is how you solve the problem.
Finally, for the question about compactness, $I = [0,1]$ and $I^\mathbb{N}$ refers to the space $\prod_{n=0}^{\infty} [0,1]$.  There is a hard theorem about compactness of products of spaces.  But here you are probably expected to use something more elementary.  So, here's a hint that will do the trick.  The continuous image of a compact space is compact.  Suppose you can show $P$ is compact.  Is $I^\mathbb{N}$ the continuous image of $P$?
