Which of the following are true in topological spaces?

Question

Let $T_1$ and $T_2$ be two topologies defined on $\mathbb{N}$ (the set of all natural numbers), where $T_1$ is the topology generated by

$$B = \{\{2n-1, 2n\}: n \in \mathbb{N}\}$$

and $T_2$ is the discrete topology on $\mathbb{N}$. Consider the following statements:

1. In $(\mathbb{N}, T_1)$ every infinite subset has a limit point.

2. The function $f: (\mathbb{N}, T_1)\to (\mathbb{N}, T_2)$ defined by $$f(n)= \begin{cases} \frac{n}{2}, \ \ n\ \text{ even}\\\frac{n+1}{2}, \ \ n\ \text{odd}\end{cases}$$ is a continuous function.

Which of the above statements are TRUE?

My attempt: Since $\{1\}$ is an open set in $T_2$, so $f^{-1}(\{1\})= 1$ which is not open in $T_1$. Am I correct? What about first option?

Answer 1

We saw in the other answer that 2 is in fact true, $f$ is continuous.

But 1 is also true: suppose $A$ is a non-empty subset of $\mathbb{N}$ (in the topology $T_1$). If $2n \in A$, then $2n-1$ is a limit point of $A$, because every open set that contains $2n-1$ contains $\{2n-1,2n\}$ and so intersects $A\setminus \{2n-1\}$. And if some odd number $2n-1$ is in $A$ then by a similar reasoning $2n$ is a limit point of $A$ as well. And $A$ must contain an odd or even number (and so has a limit point).

So we have here an example of a limit point compact (also called weakly limit point compact in some books) space $(X,T_1)$ and a continuous function from $(X,T_1)$ onto $(X,T_2)$ where the last space is clearly not limit point compact at all, showing that this weak compactlike property is not always preserved under continuous images (unlike compactness, sequential compactness, Lindelöfness, strong limit point compactness and other such properties).

Answer 2

You have that for your topological space $(\mathbb{N},T_2)$, the set $$\mathcal{B}_2=\{\{n\}\mid n\in\mathbb{N}\}$$ i.e. the set of singletons in $\mathbb N$, forms a basis.

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Note second, that for two topological spaces $(X,\mathcal T)$, $(Y,\mathcal S)$ with a basis $\mathcal B$ for $\mathcal S$, if a function $g:X\to Y$ satisfies that $g^{-1}(A)\in\mathcal T$ for every $A\in\mathcal B$, then $g$ is already continuous.

To see this, let $S\in\mathcal S$. Then $S=\bigcup\mathcal A$ for some $\mathcal A\subseteq\mathcal B$. Then $$f^{-1}[S]=f^{-1}[\bigcup \mathcal{A}] = \bigcup \{f^{-1}[A]: A \in \mathcal{A}\} \in \mathcal{T}$$ as it was supposed that $f^{-1}(A) \in \mathcal{T}$ for all $A \in \mathcal{B}$.

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Thus, we just have to verify that $f^{-1}(\{n\})\in T_1$ for every $n\in\mathbb N$.

This however follows as if $f(x)=n$ for $x\in\mathbb N$, then either $x$ is even or odd. If $x$ is even, then $n=f(x)=x/2$, i.e. $x=2n$. If $x$ is odd, then $n=f(x)=(x+1)/2$, i.e. $x=2n-1$. Thus only $2n$ and $2n-1$ could map to $n$ over $f$ and you can verify that they also do map to $n$ over $f$.

Thus $f^{-1}(\{n\})=\{2n,2n-1\}\in B\subseteq T_1$ as of your definition for $B$.

So 2. is true.