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Let $X$ be infinite set with cocountable topology

Let Y be topological space and $f:X \to Y$ and $x_n \to x$ show $f(x_n)=f(x)$

extra: does this mean it is countinous

Defintions

  • Given an uncountable set $X$, the co-countable topology on $X$ is $$T_{\text{co-countable}} =\{ A \subset X : A^c \cap X \text{ is countable }\} \cup \{ \emptyset \}.$$

  • Let $(X,T)$ be a topological space. If $(x_n)_{n=1}^\infty$ is a sequence in $X$ we say it converges to $x_0 \in X$, $x_n \to x_0$,

    if $\forall $ open set $U \subset X$ s.t $x_0 \in U$, $\exists$ $n_0 \in \mathbb N$ s.t $x_n \in U$ for all $n \geq n_0$.

Def of $f(x_n) =f(x)$

  • Let $(Y,T_2)$ be a topological space. If $(f(x_n))_{n=1}^\infty$ is a sequence in $Y$ we say it converges to $f(x) \in X$, $f(x_n) \to f(x_0)$,

    if $\forall $ open set $V \subset Y$ s.t $f(x_0) \in V$, $\exists$ $n_0 \in \mathbb N$ s.t $f(x_n) \in V$ for all $n \geq n_0$.

missing of defigion of mapping in a topological spaece

I am trying to tied it toghter like its real. but dont see it

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    Hint: which sequences converge in the cocountable topology? – 2017-02-16
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    Also the question as stated is unclear, you have $x_n\to x$ and then you use $f(y)$ without having defined $y$ – 2017-02-16
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    constant, or alternating as long as thier not infintely different. or it might be for another topology – 2017-02-16
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    regardless of $f$, if $x_n\to x$ then you will have that $x_n=x$ for all but finitely many $n$. Therefore, for any $f$, continuous or not, you will have $f(x_n)=f(x)$ for all but finitely many $n$. (If there are infinitely many $n$ for which $x_n\not=x$ then throw these $x_n$ out, you get a neighborhood $U$ of $x$ which shows you that $x_n\not\to x$, a contradiction.) – 2017-02-16

2 Answers 2

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Let $(x_n)$ be a sequence in $X$ in the co-countable topology, that converges to $x$. I claim that the sequence is eventually constantly $x$: $\exists n_0: \forall n \ge n_0: x_n = x$

To this end define $N = \{x_n: n \in N\} \setminus \{x\}$. This set is at most countable and $x \notin N$, so $U = X\setminus N$ is open in the co-countable topology and $x \in U$. Then by the definition of convergence $\exists n_0: \forall n \ge n_0: x_n \in U$. This $n_0$ is as required as $x_n \in U$ implies $x_n = x$ (all $x_n \neq x$ are in $N$, so not in $U$).

So this implies that if $x_n \rightarrow x$ ,then $f(x_n) \rightarrow f(x)$, as the image of an eventually constant-$x$ sequence is also eventually constant-$f(x)$ hence convergent to $f(x)$ in any topology.

Conclusion: any function defined on the co-countable topology is sequentially continuous, but not necessarily continuous. Consider the identity to a space with a finer topology: so $f(x) =x $ from $X$, cocountable to $X$ discrete (for uncountable $X$) is not continuous but it is sequentially continuous.

This is analogous to the fact that any function on a discrete space is continuous.

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Define $N=\lbrace x_n\rbrace_{n\in\mathbb{N}}$ and suppose $x\notin N$. Then taking $U=X\setminus N,x\in U\in T$ and $\forall n\in\mathbb{N},x_n\notin U$ so $x_n\not\rightarrow x$. Therefore $x\in N$, so $\exists n_0\in\mathbb{N},x=x_{n_0}$, so $f(x)=f(x_{n_0})$.

The question is unclear so I don't know if until this point I answered it.

If the answer is what you searched, then $f$ isn't necessary continuous.

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    . I'm sure it's right. I don't under why the first sentence – 2017-02-16
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    I can't follow the argument. Is this contradiction? Why doesn't xn to x? – 2017-02-16
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    @Ugo I supposed $x\notin N$ and get a contradiction so proved $x\in N$. The argument was $\exists U\in\tau_x,\forall n\in\mathbb{N},\exists m_n,m_n\geq n:x_{m_n}\notin U$ (in this case $m_n=n$) which is the negation of $x_n\rightarrow x$ in topology terms – 2017-02-16