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I've thought about this question for a while, from Basener's Topology and its Applications.

Let $X$ be a metric space with a metric $d$ and $Y \subseteq X$. Prove that the subspace topology on $Y$ inherited from $X$ is the same as the metric topology from the metric $d$ on $Y$.

Here's an attempt that seems fuzzy: Choose $U \subset Y$ open. Then $U$ is open in $X$ under the subspace topology. But one can construct a ball $B(x)$ with some radius $r$, where $x \in U$ using the metric $d$, and so $U$ is open in $X$ in the $d$-metric topology.

My thought is that two topologies are the same if the open sets of one are also open in the other. I am unsure, however, if my approach is correct.

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  • $\begingroup$ Your thought is correct. You need the same open sets in either topology. However, one correction. $U$ open in $Y$ in the subspace topology doesn't necessarily mean $U$ open in $X$. Why is that? $\endgroup$
    – user403337
    Dec 30, 2019 at 4:04
  • $\begingroup$ Looking at what you wrote again, it is a little fuzzy. You seem to have interchanged $X$ and $Y$. That is, $Y$ is supposed to have the subspace topology. $X$ contains $Y$. $\endgroup$
    – user403337
    Dec 30, 2019 at 4:12
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    $\begingroup$ I still don't quite follow. From my understanding, if $(X, T)$ is a topological space, then giving $Y$ the subspace topology means that for a set $S \subset Y$, $S$ is open in $Y$ if $S = U \cap Y$ for open $U \subset X$ $\endgroup$
    – Sean Roberson
    Dec 30, 2019 at 4:37
  • $\begingroup$ That's correct. But it follows that $S$ needn't be open in $X$. Suppose $Y$ isn't open in $X$. As an example, $Y$ is always open in $Y$. $\endgroup$
    – user403337
    Dec 30, 2019 at 4:44
  • $\begingroup$ You cant say that U is open in X just because U is a subset of Y. That holds when Y is open in X. You need to show that the two topologies are equal. $\endgroup$
    – user643073
    Dec 30, 2019 at 10:14

2 Answers 2

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If $d_Y$ is the metric of $X$ restricted to $Y$ (so essentially $d\restriction_{Y \times Y}$), the main observation is that for the balls in this metric we have, almost by definitions:

$$\forall y \in Y: \forall r>0: B_{d_Y}(y,r) = B_d(y,r) \cap Y$$

so that the $d_Y$-balls in $Y$ are relatively open in $Y$, which already implies that the metric topology on $Y$ induced by $d_Y$ is a subset of the subspace topology, and on the other hand if $O$ is open in $X$, and $y \in O \cap Y$, some $d$-open ball around $y$ sits inside $O$ and then the intersected ball (thus a $d_Y$ ball by the above) sits inside $O \cap Y$, making all members of $O \cap Y$ interior points in the $d_Y$-topology, and so the subspace topology is a subset of the topology induced by $d_Y$. Hence the equality of topologies.

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Let $U$ be open within $Y$.
Thus there exists $V$ open within $X$ with $U = V \cap Y$.
For all $x$ in $V$, exists ball $B(x)$ with $B(x) \subseteq V$.
$V = \bigcup \{ B(x) : x \in V \}$.

$$U = V \cap Y = \bigcup \{ B(x) : x \in V \} \cap Y = \bigcup \{ B(x) \cap Y : x \in V \} = \bigcup \{ B(x) \cap Y : x \in X \}$$

(The last equality requires some thought to prove.)
As every open set of $Y$ is the union of a collection of balls of $Y$,
$Y$ is metrized by the metric of $X$.

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  • $\begingroup$ Last equality: It's enough to intersect with open balls centered in $X$ as long as they intersect with $Y$. $\endgroup$
    – Sean Roberson
    Dec 30, 2019 at 7:32
  • $\begingroup$ @SeanRoberson. The topology of z $\endgroup$
    – William Elliot
    Dec 30, 2019 at 8:18
  • $\begingroup$ The topology of a metric space is the topology of all the open balls centered at every point in the space. It does not include balls centered in alien spaces. The balls that generate the topology of the real line do not include balls centered at a + bi, b /= p. $\endgroup$
    – William Elliot
    Dec 30, 2019 at 8:33

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