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The cofi nite topology on $\mathbb{R}$ is the topology in which a subset $F⊆\mathbb{R}$ is closed if and only if $F$ is either finite or $F = \mathbb{R}$. Let $X = \mathbb{R}$ with the co nite topology and $Y = \mathbb{R}$ with the usual topology. Show that any continuous map $f : X →Y$ is a constant.

I was trying to solve this problem by contradiction but could proceed much. can anyone help me please to tackle this problem

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HINT: Suppose that $f:X\to Y$ is not constant; then there are $x_0,x_1\in X$ such that $f(x_0)\ne f(x_1)$. Let $U_0$ and $U_1$ be disjoint open nbhds of $f(x_0)$ and $f(x_1)$, respectively. Then $f^{-1}[U_0]$ and $f^{-1}[U_1]$ are open nbhds of $x_0$ and $x_1$ in $X$. Use the fact that $U_0\cap U_1=\varnothing$ to get a contradiction.

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I was tackling the problem same way as you talk. since $X$ is not Hausdorff I was thinking that $U_0\cap U_1=\varnothing$ holds or not. – user59908 Mar 14 '13 at 15:28
@user59908: I think that you mean $f^{-1}[U_0]$ and $f^{-1}[U_1]$. The fact that $X$ is not Hausdorff is not by itself enough to guarantee that $f^{-1}[U_0]\cap f^{-1}[U_1]\ne\varnothing$, but the definition of the cofinite topology is: try proving that if $V$ and $W$ are non-empty open sets in $X$, then $V\cap W\ne\varnothing$. – Brian M. Scott Mar 14 '13 at 15:32
sorry printing mistake for me . – user59908 Mar 14 '13 at 15:41
I was not sure how to proceed to a contradiction from the fact that $U_0\cap U_1=\varnothing$ – user59908 Mar 14 '13 at 15:44
@user59908: Since $U_0\cap U_1=\varnothing$, you know that $f^{-1}[U_0]\cap f^{-1}[U_1]=\varnothing$ also. Now show that this is impossible, by showing that if $V$ and $W$ are non-empty open sets in $X$, then $V\cap W\ne varnothing$. It will help to remember that the union of two finite sets is still finite. – Brian M. Scott Mar 14 '13 at 15:46

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