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Question (from my intro to topology course): Let $ (X, \mathcal{T}) $ be a metrizable topological space and let $ N $ be a neighborhood of a point $ a \in X $. (A neighborhood $ N $ of $ a $ of a topological space $ (X, \mathcal{T}) $ is a subset of $ X $ that contains an open set that contains $ a $.) Prove that $ N $ contains a neighborhood $ V $ of $ a $ such that $ V $ is closed.

I just started to learn about topological space after learning about metric space and I am confused about the definitions of neighborhood and open set. This thread explains that I can define open sets in a topological space $ (X, \mathcal{T}) $ to mean elements in the topology $ \mathcal{T} $ or open sets in a metric space $ (X, d) $ to mean a set that is a neighborhood of each of its points.

Going back to my original problem, if $ (X, \mathcal{T}) $ is metrizable, then I can define a metric $ d $ on $ X $ whose open sets form the topology $ \mathcal{T} $. Since $ N $ is a neighborhood of $ a \in X, N $ contains an open set, say $ O_{a} $, that contains $ a $. Is $ O_{a} $ in this case an element of $ \mathcal{T} $ or is it an open set in the metric space $ (X, d) $? How can I prove that $ N $ contains a neighborhood $ V $ of $ a $ such that $ V $ is closed? My current approach is to find an open set $ V $ such that $ N^{C} \subset V $, but haven't succeeded so far. For this current approach, is the open set $ V $ that I need to find defined as an element of $ \mathcal{T} $ or as an open set in the metric space $ (X, d) $?

Any insight is appreciated.

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If $N$ Is a neighbourhood of $x$ and $d$ is a compatible metric, for some $ r >0$, we know that $x \in B_d(x,r) \subseteq N$. But then $V:= D_d(x,\frac{r}{2}) =\{y: d(x,y) \le \frac{r}{2}\}$ is a closed neighbourhood of $x$ inside $B_d(x,r)$ hence inside $N$. That $D_d(x,s)$ is closed for any $s>0$, is standard in metric topology, and it's a neighbourhood of $x$ as $B_d(x, \frac{r}{2})$ is contained in it.

Henno Brandsma
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  • When you define the neighborhood $ N $ in that way, you are considering $ N $ as a neighborhood in a metric space $ (X, d) $ right? (A neighborhood $ N $ of $ x $ in a metric space $ (X, d) $ contains an open ball $ B(x, r) $ for some $ r > 0 $.) Also, why is $ D_{d}(x, r) $ closed? $ D_{d}(x, r) = { y : d(x, y) < r/2 } = B_{d}(x, r/2) $, so should it be open instead? – newbie Mar 05 '17 at 18:57
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    $N$ is a neighbourhood in a metrisable space, so it contains a ball. $D_d$ has $\le$ as I wrote. $(X, d)$ is just the same as $(X, \mathcal{T})$. – Henno Brandsma Mar 05 '17 at 18:58