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Consider two topological spaces:

  1. The set of positive integers $\mathbb{N}$ and the topology $\mathscr{M}$ given by $\emptyset$ and the subsets of $\mathbb{N}$ that contain $1\in\mathbb{N}$.
  2. The set $[0,1]\in\mathbb{R}$ with the metric $d(t,s)=|t-s|$.

I want to determine íf the function $f:[0,1]\rightarrow \mathbb N$ defined by $$ f(t):= \begin{cases}n, & t=0 \\ 1, & t \in] 0,1[ \\ m, & t=1\end{cases} $$ is continuous. I know that $f$ is continuous if $$\forall t\in[0,1] \text{ and }\forall U\in\mathfrak{U}(f(t)) :f^{-1}(U)\in\mathfrak{U}(t). $$ So if I can show that $f^{-1}(U)\in\mathfrak{U}(t)$ for any $U\in\mathfrak{U}(n)\cup\mathfrak{U}(1)\cup\mathfrak{U}(m)$ then I can conclude that $f$ is continuous. For $t=0$, the neighborhoods $U\in\mathfrak{U}(f(t)) $ are sets that contain $n$ and $1$, so $[0,1)\subseteq f^{-1}(U)$. This is a neighborhood of $t=0$ if there exists an open ball in $[0,1)$ that contains $0$. However, does this open ball exist when the set is closed from the bottom? And am I approaching the problem of determining if $f$ is continuous correctly?

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1 Answer 1

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Yes, you are approaching it correctly. Note that, in $[0,1]$, $[0,1)$ is an open set, since it is the ball centered at $0$ with radius $1$.

By a similar argument, $(0,1]$ is an open subset of $[0,1]$, and therefore $f$ is continuous at $1$ too.

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  • $\begingroup$ Great, thanks a lot. $\endgroup$
    – Hydrogen
    May 22 at 17:05
  • $\begingroup$ I'm glad I could help. $\endgroup$ May 22 at 17:19

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