Ytukyg

Here is Prob. 1, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:

Give $[0, 1]^omega$ the uniform topology. Find an infinite subset of this space that has no limit point.

Here is a Mathematics Stack Exchange on this very problem.

My Attempt:

We note that $[0,1]^omega$ denotes the set of all the sequences $left( x_n right)_{n in mathbb{N}}$ of the real numbers $x_n$ in the closed unit interval $[0,1]$, and that the uniform metric $bar{rho}$ on $[0,1]^omega$ be given by
$$
bar{rho}left( left( x_n right)_{ninmathbb{N}} , left( y_n right)_{ninmathbb{N}} right) colon= sup left{ min left{ leftvert x_n - y_n rightvert, 1 right} colon n in mathbb{N} right} tag{A} $$
for all $left(x_nright)_{n in mathbb{N} }, left(y_n right)_{ n in mathbb{N} } in [0,1]^omega$.

We show that $[0,1]^omega$ is not limit-point-compact

Let
$$A colon= left{ (x_n)_{ninmathbb{N}} colon x_n in { 0 , 1 } mbox{ for all } n in mathbb{N} right} = { 0, 1 }^omega. tag{B} $$
This set $A$ is an infinite subset of $[0, 1]^omega$.

We show that the set $A$ has no limit points in $[0,1]^omega$.

Let $ mathbf{x} colon= (x_n)_{ninmathbb{N}}, mathbf{y} colon= (y_n)_{ninmathbb{N}} in A$ be any two distinct points. Then
$x_n, y_n in { 0, 1}$ for each $n in mathbb{N}$ and $x_k neq y_k$ for at least one $k in mathbb{N}$, and for this $k$, we have $x_k - y_k = pm 1$. Therefore we find that
$$ bar{rho} left( mathbf{x}, mathbf{y} right) geq leftlvert x_k - y_k rightrvert = 1, $$
but we know from (A) above that
$$ bar{rho} left( mathbf{x}, mathbf{y} right) leq 1; $$ therefore
$$ bar{rho} ( mathbf{x}, mathbf{y} ) = bar{rho}left( (x_n)_{ninmathbb{N}}, (y_n)_{ninmathbb{N}} right) = 1. tag{1} $$
for any two distinct points $mathbf{x}, mathbf{y} in A$.

Let $mathbf{p} colon= left( p_n right)_{n in mathbb{N} }$ be any point of $[0, 1]^omega$. We now show that this point $mathbf{p}$ cannot be a limit point of set $A$.

Case 1. Suppose that $mathbf{p} in A$. Let us take any real number $epsilon$ such that $0 < epsilon < 1$. Then for any point $mathbf{a} in A$ such that $mathbf{a} neq mathbf{p}$, we must have
$$ bar{rho} ( mathbf{a}, mathbf{p} ) = 1 > epsilon, $$
because of (1) above. Therefore
$$ left( A cap B_{bar{rho}} ( mathbf{p}, epsilon ) right) setminus { mathbf{p} } = emptyset. $$
So $mathbf{p}$ is not a limit point of $A$.

Case 2. Suppose that $mathbf{p} in [0, 1]^omega setminus A$. Then there exists at least one $N in mathbb{N}$ for which $p_N notin { 0, 1}$; for that $N$ we have $0 < p_N < 1$. Now let us take any real number $epsilon$ such that
$$ 0 < epsilon < min left{ frac{p_N}{2}, frac{1-p_N}{2} right}. $$
Then
$$ epsilon < frac{p_N}{2} qquad mbox{ and } qquad epsilon < frac{1-p_N}{2}, $$
and so
$$ - frac{p_N}{2} < - epsilon qquad mbox{ and } qquad epsilon < frac{1-p_N}{2}, $$
which in turn implies that
$$ frac{p_N}{2} = p_N - frac{p_N}{2} < p_N - epsilon qquad mbox{ and } qquad p_N + epsilon < p_N + frac{1 - p_N}{2} = frac{p_N + 1}{2}. $$
In fact we obtain
$$ 0 < frac{p_N}{2} < p_N - epsilon qquad mbox{ and } qquad p_N + epsilon < frac{p_N + 1}{2} < 1. tag{2}$$
Now if $mathbf{x} colon= left( x_n right)_{n in mathbb{N} }$ in $[0, 1]^omega$ satisfies
$$ bar{rho}( mathbf{x}, mathbf{p} ) < epsilon, $$
then for that point $mathbf{x}$ we also have
$$ leftlvert x_N - p_N rightrvert < epsilon, $$
which implies that
$$ p_N - epsilon < x_N < p_N + epsilon, $$
which together with (2) above implies that
$$ 0 < frac{p_N}{2} < p_N - epsilon < x_N < p_N + epsilon < frac{p_N + 1}{2} < 1, $$
and thus
$$ 0 < x_N < 1, $$
from which it follows that $mathbf{x} notin A$. Thus we can conclude that
$$ A cap B_{bar{rho}}( mathbf{p}, epsilon ) = emptyset, tag{3}$$
and so
$$ left( A cap B_{bar{rho}}( mathbf{p}, epsilon )right) setminus { mathbf{p} } = emptyset. $$
Thus $mathbf{p}$ is again not a limit point of our set $A$. In fact, $mathbf{p}$ is not even an adherent point of set $A$.

From (3) we can even show that our set $A$ is in fact a closed set in $[0, 1]^omega$.

Am I right?

Finally, it is my understanding that the above proof can go through even if we take our set $A$ as follows: Let $alpha$ and $beta$ be any two distinct real numbers such that $alpha in [0, 1]$ and $beta in [0, 1]$. Then let us put
$$ A colon= left{ left( x_n right)_{n in mathbb{N}} colon x_n in { alpha, beta } mbox{ for all } n in mathbb{N} right} = { alpha, beta }^omega. tag{B*} $$

Is my proof correct and clear in all its details? If not, then where is it in need of improvement?