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Let $X$ be a $T_1$-space, and show that $X$ is normal if and only if each neighbourhood of a closed set $F$ contains the closure of some neighbourhood of $F$.

I saw this statement in the book Introduction to topology and modern analysis by Simmons. I tried to prove it but I did not get any where. How may it be proved?

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So you want to show the following: $X$ normal is equivalent to

(*) for all closed subsets $F$ of $X$ and every open set $U$ with $F \subseteq U$ there exist some open set $V$ such that $F \subseteq V \subseteq \overline{V} \subseteq U$.

To show(*) from normality suppose we have $F \subseteq V$ as in the premise.

Then $F$ and $G := X\setminus V$ are closed and disjoint. So by normality we have open subsets $O_1$ and $O_2$ that are disjoint and such that $F \subseteq O_1$ and $X \setminus V \subseteq O_2$ The latter means that $X \setminus O_2 \subseteq V$ (And the former set is closed) and also $O_1 \subseteq X\setminus O_2$ by disjointness, we get

$$F \subseteq O_1 \subseteq \overline{O_1} \subseteq \overline{X \setminus O_2} = X \setminus O_2 \subseteq V\text{,}$$

so we can take $U = O_1$.

On the other hand if ($\ast$) holds and $F, G$ are closed and disjoint, apply (*) to $F$ and $U = X \setminus G$, and take $O_1 = V$ and $O_2 = X\setminus \overline{V}$ to get disjoint open neighbourhoods of $F$ and $G$.

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    thank you very much! i understood the first half, but i did not get the idea of the second half, how i may apply * to X-G once it is open, also what guarantees G to be included in O_2? – 2017-01-24
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    @YassinRany I made a typo, so I made it clearer I hope – 2017-01-24
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    Thank you again! you saved my day haha, just going more far if i may, you seems to have a long experience with topology, i am just a beginner, these kind of prove though looking intuitive as whole, do not seems something intuitive to me, there is a common idea of proof theorems like this, i mean it is a ``traditional way`` or some particular idea and development ? – 2017-01-24
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    @YassinRany For me, this proof is completely obvious, as soon as you realise the idea of taking complements. It's indeed a standard idea. Often notions that are formulated in terms of open sets can be translated to an equivalent one with closed sets, and vice versa. – 2017-01-24
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    Really thank you for the attention and help ))) – 2017-01-24