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Let $X = \mathbb R$ and $Y = \{y_1,y_2,\dots\}$ be a bounded subset of $X$. Is every subset of $Y$ an open set in the subspace topology of $Y$ induced by $X$?

More concretely, I am considering the set $\{1/n : n \in \mathbb N\}$ as a subset of $\mathbb R$. Every open set in the subspace topology is the restriction of open sets in $\mathbb R$ to $Y$, so can't I cover each point in any subset of $Y$ with some open interval?

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  • $\begingroup$ In general it is false. What topology in $\mathbb{R}$ are you using? $\endgroup$ – sinbadh Jan 27 '16 at 7:34
  • $\begingroup$ Does anyone use anything other than the standard topology on $\R$? The other ones are so useless... $\endgroup$ – Schmidt Jan 29 '16 at 6:57
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I am interpreting your question as "is every countable bounded subset of $\Bbb R$ a discrete space under the subspace topology?", in which case the answer is no.

$A = \{0\} \cup \{1/n| n \in \Bbb N\} \subset \Bbb R$ is countable and bounded. Yet, as every neighborhood of $0$ in $\Bbb R$ contains all but finitely many numbers of the form $1/n$, $\{0\}$ is not an open set of $A$. More simply, $\Bbb Q \cap [0, 1] \subset \Bbb R$ is not a discrete space under the subspace topology, since rationals are dense.

However, $\{1/n|n \in \Bbb N\} \subset \Bbb R$ (i.e., $A$ without it's limit point $0$) indeed inherits the discrete topology. For given any $1/n$ in $\Bbb R$, one can pick a small neighborhood around it which misses the other points.

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