What is the "topology induced from $\mathbb R^2$?

Question

Consider the set $S$ of all the points in the plane of coordinates $(x,sin(1/x))$ for $x \in (0,1]$. Then, take the closure of it ${S'}$ and show that this closed set, viewed as a metric space with the topology induced from $\mathbb R^2$, is connected but not arcwise connected.

What is the "topology induced from $\mathbb R^2$"? I am taking analysis right now and so we barely covered topology so please explain in the context of metric spaces. BTW, what are the limit points? Is it $(0,y)$, where $y \in \{0,1,-1\}$?

The topology induced from $\Bbb R^2$ is the subspace topology, i.e. $S\subseteq (X,\tau)$ has the subspace topology $\tau_S$ when $A\in \tau_S\Leftrightarrow \exists B\in \tau, A=S\cap B$. If $X$ is a metric space, then the subspace topology is the topology induced on $S$ by the restriction of the metric to $S\times S$. — Sassatelli Giulio, May 17 '23 at 15:49
@RobertShore Is the only limit point $(0,0)$? Is $(0,-1), (0,1)$ limit points? and hence part of the closure? — beginner, May 17 '23 at 16:44

score 1 · Answer 1 · answered May 17 '23 at 15:50

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The induced topology on $S'$ is just the subspace topology it inherits from $\mathbb R^2$.

That is, open sets in $S'$ are of the form $U\cap S'$, where $U$ is open in $\mathbb R^2$.

answered May 17 '23 at 15:50

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Okay, I remember covering this actually very briefly though my intuition isn't great. How do you recommend I proceed / what should I think about? – beginner May 17 '23 at 16:07

What is the "topology induced from $\mathbb R^2$?

1 Answers1