Let $X_n = \mathbb{R}$ for all $n \in \mathbb{N}.$ Now set $Y = \Pi_{n \in \mathbb{N}} X_n.$ Endow $Y$ with the product topology, where the topology on each $X_n$ is the Euclidean topology on $\mathbb{R}.$
Consider the subset $S$ of $Y$ defined by $$S=\{ (x,x,\cdots )| x \in \mathbb{R}\},$$ with the subspace topology induced from $Y.$
Comment on the following properties of the space $S.$
(i) $S$ is open in Y.
(ii) $S$ is dense in Y.
(iii) $S$ is disconnected.
(iv) $S$ is locally compact.
Attempt: Construct a map $f:S \rightarrow \mathbb{R}$ defined by $$f(x,x,\cdots)=x.$$ Since $f$ is a homeomorphism, we have that $S$ is both locally compact and connected as $\mathbb{R}$ is both locally compact and connected.
But I am stuck on how to approach (i) and (ii).
Please help.
$S=\bigcap_n \{(x_k): x_n=x_{n+1}\}$ so $S$ is closed in $Y$. [Recall that projection maps are continuous for the product topology]. This rules out (ii).
Also, $Y$ is connected so $S$ cannot be both open and closed. So (i) is also ruled out.
Alternatively, $(0,0,0,...)\in S$ and any neighborhood of this point contains $(-\epsilon, \epsilon)\times \mathbb R \times \mathbb R\times...$ for some $\epsilon >0$. So no neighborhood of $(0,0,0,...)$ is contained in $S$
Closedness of $S$: Fix $n$. Let $S_n=\{(x_i): x_n \neq x_{n+1}\}$. Suppose $(x_i) \notin S_n$. Let $r=|x_n-x_{n+1}|$ so that $r>0$. Consider $U=\{(y_i): |y_i-x_i| <r/2 \, \text {for} 1\leq i \leq n+1\}$. Then $U$ is an open set in $Y$ containing $(x_i)$ which is contained in $Y\setminus S_n$. Hence, $Y\setminus S_n$ is open and $S_n$ is closed.