# Visualization / sketch for this basic proof about subspace topology

Let $$(X,d)$$ be a metric space and $$A\subset X$$ a subset equipped with the induced metric $$d_{A}$$. Then the open subsets of $$(A,d_{A})$$ are exactly the intersections of open subsets of of $$(X,d)$$ with $$A$$: $$B \subset A$$ is open in $$(A,d_{A})$$ iff there exists an open subset $$Y \subset X$$, so that $$B = A \cap Y$$.

My proof of "$$\implies$$" is pretty straight forward:

Let $$B\subset A$$ be an open subset. Define $$\begin{equation*} U_{\varepsilon}^{Z}(x) := \{ y \in Z: d_A(x,y) = d(x,y) < \varepsilon\} \end{equation*}$$ for any subset $$Z \in X$$. Because $$B$$ is open, for every $$x \in B$$ there exists an $$\varepsilon_{x} > 0$$ so that $$\begin{equation*} B\supset U^{A}_{\varepsilon_{x}}(x) = A\cap U^{X}_{\varepsilon_{x}}(x). \end{equation*}$$ The set $$\begin{equation*} Y := \bigcup_{x \in B}U^{X}_{\varepsilon_{x}}(x) \end{equation*}$$ is open in $$X$$ because it's the union of open sets in $$X$$. Furthermore we have $$B = A\cap Y$$.

But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $$U^{A}_{\varepsilon_{x}}(x) \neq U^{X}_{\varepsilon_{x}}(x)$$, since when I draw $$X$$ to be a box and $$A$$ a circle inside it and $$B$$ a smaller circle in $$B$$, we alway have $$U^{A}_{\varepsilon_{x}}(x) = U^{X}_{\varepsilon_{x}}(x)$$.

Any help is greatly appreciated :)

Draw $$X$$ to be the plane and $$A$$ to be the $$x$$-axis.
• This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis? – Viktor Glombik Mar 8 at 18:12
• Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis. – Lee Mosher Mar 8 at 19:14
• Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $\mathbb R$. – Lee Mosher Mar 8 at 21:17