Are either of these Subsets of $\mathbb{R}^{\infty}$ Compact Define a set of $\infty\times n$ matrices by
$$X_0=\lbrace [a_{i,j}]:\sum_{i,j}a_{i,j}=1, a_{i,j}\ge0,\text{ each column has positive sum}\rbrace$$
where $i\le 1$, $1\le j \le n$, and the sums are over all $i,j$. Are either of these subsets of $\infty\times n$ matrices compact? 
Since these are subsets of $\mathbb{R}^{\infty}$, the "compact iff closed and bounded" doesn't apply here. I am having difficulty applying the open covering definition of compactness due to the fact that I have infinitely many rows.
Edit: I have a mapping $T=f\circ g:X_0\to X_0$, where $g$ multiplies each column by a constant to obtain column sum $1/n$ (so that the sum of all the columns is $1$) and then $f$ multiplies each row by a constant to obtain some desired row sum $r_i$ (such that $\sum_i r_i=1$). Which topology can I place on this set $X_0$ such that $X_0$ is compact and $T$ is continuous? If $X,Y\in X_0$, I believe I can show that $X\to Y$ (meaning $x_{i,j}\to y_{i,j}$) implies $T(X)\to T(Y)$, but I am unsure of which topology is being implicitly used when I conclude this gives continuity. Perhaps it's the product topology.
 A: My attempt:
Take the topology on $(\mathbb{R}^{d})^{\mathbb{N}}$ to be the product topology. Suppose $d = 2$, let $f_n \in (\mathbb{R}^{d})^{\mathbb{N}}$.  be such that $f_n(1,1) = \frac{1}{n}$, $f_n(1,2) = \frac{n-1}{n}$, $f_n = 0$ elsewhere then $f_n \in X_0$, however $f = \lim_n f_n$ is such that $f(1,1) = 0$, the first column sum is $0$ hence $f \not\in X_0$, $X_0$ is not closed so a priori not compact since $(\mathbb{R}^{d})^{\mathbb{N}}$ is Hausdorff. One can easily extend this to $d > 2$.
For $d = 1$, take $f_n(k) = n^{-1}$ for $k \leq n$, $0$ otherwise. One can also take $f_n(k) = \frac{1}{n} (1 - n^{-1})^{k-1}$. In either case $f_n \in X_0$, but $f_n \to 0$.
Hence in any case $X_0$ fails to be compact.
If you insist on the column sums to be strictly positive, I do not think it is likely that $X_0$ is compact in any reasonable topology. On the other hand let $X'_0 = \{ \sum_{i} {a_{i}} = 1, a_{i} \geq 0 \}$, you can take $X'_0$ as a subset of $l_{\infty}$ and it is $\sigma(l_\infty, l_1)$-compact (Banach-Alaoglu), with similar extension for $d > 1$ (it fails however to be $l_\infty$-compact).
