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Let $\mathbb {R}_S$ be defined as the standard topology whose basis elements are of the form, $\{(a,b) : a,b \in \mathbb {R}, a<b\}$ and $\mathbb {R}_K$ the $K$-topology whose basis elements are of the form, $\{(a,b) : a,b \in \mathbb {R}, a<b\} \cup \{(c,d)-K : c,d \in \mathbb {R}, a<b\}$ where $K = \{1/k : k\in\mathbb {Z}+ \}$.

Since $\mathbb {R}_S \subset \mathbb {R}_K$, shouldn't it make sense that all open sets in $\mathbb {R}_S$ also be open sets in $\mathbb {R}_K$?

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Yes.

Given two topologies on a set $X$, if one topology (call it the smaller one) is a subset of the other (call it the larger one), then since (by definition), the topology on a space is the set of open subsets of the space, any subset of $X$ which is open in the smaller topology is still open in the larger one.

Hence if the basis elements of the smaller topology are open in the larger one, then any subset of $X$ which is open in the smaller topology (being a union of basis elements of the smaller topology) is still open in the larger one.

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