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.

  • $\begingroup$ This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis? $\endgroup$ – Viktor Glombik Mar 8 at 18:12
  • $\begingroup$ 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. $\endgroup$ – Lee Mosher Mar 8 at 19:14
  • $\begingroup$ Sure, but when I want to draw a sketch, an interval is more convenient, right? $\endgroup$ – Viktor Glombik Mar 8 at 19:56
  • $\begingroup$ Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $\mathbb R$. $\endgroup$ – Lee Mosher Mar 8 at 21:17

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