Neighborhoods, convergence of quotient topology $[0,1] x N$

37 Views Asked by At

I am having a bit of difficulty wrapping my head around this:

Consider the space $$X = [0,1] \times \mathbb{N}$$ as a subspace of $\mathbb{R}^2$. Define the equivalence relation $$(x,y) \sim (z,w) \iff (x,y) = (z,w)$$ or $$x=z=0$$. Prove that the sequence $$\big[\big(\frac{1}{n},n\big)\big]_\sim$$ doesn't converge in the quotient space $X/\sim$.

To show that this does not converge, I need to find a neighborhood of $(0,n)$ which does not contain the sequence for infinitely many points. My question is, what would this neighborhood look like? Would $$\pi^{-1}\big(\big[0,\frac{1}{n+1}\big) \big)$$ work?

Thank you.

1

There are 1 best solutions below

1
On BEST ANSWER

What would that $n$ be? A fixed one? A variable one?

You don't need just an open set around the point $(0,n)$, you need an open set around the set $\{0\}\times\mathbb{N}$; that will give you a neighbourhood of the point $\{0\}\times\mathbb{N}$ in the quotient space. Try $\bigcup_n([0,\frac1{2n})\times\{n\})$.