A space $X$ called $P$-space if every $G_\delta$ subset of $X$ is open in $X$. Every discrete space, I believe, is $P$-space. My question is this: Could someone offer some other classical examples of $P$-space? Thanks ahead:)

This question already has an answer here: Arcwise connected part of $\mathbb R^2$ 3 answers If $A\subset\mathbb{R^2}$ is countable, is $\mathbb{R^2}\setminus A$ path connected? [duplicate] 1 answer Prove that the complement of $\mathbb{Q} \times \mathbb{Q}$. in the plane $\mathbb{R}^2$ is connected. [duplicate] 5 answers

I know $R^\omega$ is the set of functions from $\omega$ to $R$. I would think $R^\infty$ as the limit of $R^n$, but isn’t that $R^\omega$? The seem to be used differently, but I can’t tell exactly how.

Let $f, g: X \rightarrow \mathbb{R}$ be continuous functions, where ($X, \tau$) is a topological space and $\mathbb{R}$ is given the standard topology. a)Show that the function $f \cdot g : X \rightarrow \mathbb{R}$,defined by $(f \cdot g)(x) = f(x)g(x)$ is continuous. b)Let $h: X \setminus \{x \in X | g(x) = 0\}\rightarrow \mathbb{R}$ be […]

I am trying to prove the following: Let $X$ be a first countable space and $x$ a member of $X$. Prove that there is a local nested basis $\{S_n\}_{n=1}^\infty$ at $x$. Since $X$ is first countable there is a countable local base $\mathcal{B}_x$ at $x$. Constructing a nested sequence of subsets of $\mathcal{B}_x$ is easy. […]

Let $\mathbb{R}^\omega$ be the countable product of $\mathbb{R}$. Make it a topological space using the box topology. Show that the sequence $\{(1/n, 1/n, ….)$ | $n \in \mathbb{Z}_+\}$ does not converge to $(0,0,…)$. I know that a given sequence $\{x_n$ | $n \in \mathbb{Z}_+\}$ such that each $x_n$ is in $X$, $\{x_n\}$ converges to x […]

If $\langle A_n : n \in \omega \rangle$ is a sequence of subsets of a set $X$, $$ \underline{Lim} A_n = \{ x \in X : \exists n_0 \in \omega \forall n \geq n_0, x \in A_n \} $$ If $\mathcal A$ is a family of subsets of a set $X$, then, $L(\mathcal A)$ denotes […]

Consider the Hilbert cube $Y = [0,1]^\mathbb{N}$. It is easy to define four classes of metrics on $Y$ for $\gamma>0$ and $\omega>1$: $$d^\gamma_{sup,pol}(x,y) = \sup_{k\geq 1} |x_k-y_k|/k^\gamma,$$ $$d^{\gamma}_{sum,pol}(x,y) = \sum_{k=1}^\infty |x_k-y_k|/k^{1+\gamma}, $$ $$d^\omega_{sup,exp}(x,y) = \sup_{k\geq 1} |x_k-y_k|/\omega^k,$$ $$d^\omega_{sum,exp}(x,y) = \sum_{k=1}^\infty |x_k-y_k|/\omega^k $$ Question, do all these metrics define the same topology? does any of this […]

It is an immediate result that a map from a first-countable space is continuous iff it is sequentially continuous. I was wondering if the converse was also true. That is, is it true that if every map from a space $X$ is continuous iff it is sequentially continuous, then $X$ must be first-countable?

Suppose we start with a topology $T_1$ of X. Is there a way to get construct a sequence of topologies $T_n$ such that $T_{n – 1} \subset T_{n}$ in which there is no finer topologies in between, also that sequence is the longest one.

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