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This is probably a really dumb question, but I really can't figure out how to show axiom (c) for the subspace topology is defined. The relevant definitions are here.

As an attempted proof, let $y\in Y$. Pick some neighborhood $N$ of $y$ and $U$ be a subset of $Y$ which contains $N$. I know that $N=X\cap N'$, where $N'$ is a neighborhood of $N$ in $X$. But I have no idea how to show that $Y=X\cap Y'$ where $Y'$ is a neighborhood of $y$ in $X$. It makes sense to do this, in say, Euclidean space, but I am not really sure how to show this general, as it is not true that $Y$ is a neighborhood of $y$ in $X$, which is what I would need by definition of the subspace topology.

I really think I'm missing something simple here.

Vasting
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1 Answers1

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I'll denote $\mathcal{N}(x)$ for the neighbourhood system at $x$ in the original space $X$ and for $Y \subseteq X$ we define, for each $y \in Y$:

$$\mathcal{N}_Y(y) = \{N \cap Y \mid N \in \mathcal{N}(y)\}$$

per the definition of subspace topology, and we want to verify (c) (the enlargement axiom) for the supposed neighbourhood system $\{\mathcal{N}_Y(y)\mid y \in Y\}$.

So we have $y \in Y$, $N = N' \cap Y \in \mathcal{N}_Y(y)$, where $N' \in \mathcal{N}(y)$ and $N \subseteq U \subseteq Y$ and we want to show that $U \in \mathcal{N}_Y(y)$. Simply consider $U' = U \cup N'$. Because $N' \in \mathcal{N}(y)$ and $U' \supseteq N'$, $U' \in \mathcal{N}(y)$ by axiom c) applied to $X$.

And $$U' \cap Y = (U \cup N') \cap Y = (U \cap Y) \cup (N' \cap Y) = U \cup N = U$$

so that indeed $U \in \mathcal{N}_Y(y)$ by definition.

Henno Brandsma
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