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The gluing lemma for closed sets states:

Given a finite closed cover $\{A_i\}$ of a topological space $X$, together with continuous maps $\{f_i : A_i \to Y\}$ into some other topological space $Y$, there exists a unique continuous map $f : X \to Y$ whose restriction to each $A_i$ is equal to $f_i$.

Question: What is a good and simple counter-example when the gluing lemma fails in the case that $\{A_i\}$ is infinite, but countably so.

My attempt: I have only been able find a silly counter-example: let $\{A_\alpha\}$ be the collection of all points of $X = [0,1] \subset \mathbb{R}$, and let $f_\alpha = 0$ for all $\alpha$ except for $\alpha = \alpha_0$, for which $A_{\alpha_0} = \{0\}$ and $f_{\alpha_0} = 1$.

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Consider the space $X=\{0\}\cup\{1/n:n\in\mathbb N\}$ with the topology induced by the inclusion $X\subseteq\mathbb R$. Pick any non-continuous function $g:X\to\mathbb R$, consider the covering $\{A_i:i\in\mathbb N_0\}$ with $A_0=\{0\}$ and $A_n=\{1/n\}$ for each $n\in\mathbb N$ —this is a countable closed covering of $X$— and define the functions $f_i=g|_{A_i}:A_i\to\mathbb R$, for each $i\in\mathbb N_0$. All the $f_i$ are continuous, and there does not exists any continuous function $f:X\to\mathbb R$ such that $f|_{A_i}=f_i$ for each $i\in\mathbb N_0$.

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Why do I need to pick a non-continuous $f : X \to \mathbb{R}$? If I define $f_n := 0$ for all $n \neq 0$, and $f_0 = 1$, this would work, wouldn't it? –  Rick Feb 2 '12 at 2:34
    
If you pick a continuous $f$ then you are not going to have a counterexample! –  Mariano Suárez-Alvarez Feb 2 '12 at 2:35
    
No, but if I define my functions $f_n$ that I want to glue together in the manner I said, then they already define a unique function $f:X \to \mathbb{R}$, simply from set-theoretic considerations. Yet, this $f$ obviously cannot be continuous. I think this does it. –  Rick Feb 2 '12 at 2:38
    
To be honest, I do not understand what you are saying in these comments... –  Mariano Suárez-Alvarez Feb 2 '12 at 2:42
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@Rick: You did pick a discontinuous $f:X\to\mathbb R$, just as Mariano suggested. Namely, your example $f(0)=1$ and $f(1/n)=0$ for all $n\neq 0$ serves as a special case of the "$g$" in this answer. Whether you decided to think of the restrictions before you think of the "glued" function doesn't affect the validity. –  Jonas Meyer Feb 2 '12 at 2:54

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