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In wiki article of "compact spaces", they state that the set $\mathbb{Q} ∩ [0,1]$ is not compact because the sets of rational numbers in the intervals $[0, \frac{1}{π} - \frac{1}{n}]$ and $[ \frac{1}{π}+ \frac{1}{n}, 1]$ covers all the rationals in $[0,1]$ but this cover does not have finite subcover; these sets are open in the subspace topology even though they are not open as subsets of $\mathbb{R}$.

I don't get it! How is $[0, \frac{1}{π} - \frac{1}{n}]$ open in subspace topology for each $n∈ \mathbb{N}$ ? In particular, I need to know what are the open sets under subspace topology? What are the closed sets in subspace topology? What are the compact sets in subspace topology?

Rushabh Mehta
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Akash Patalwanshi
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2 Answers2

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If $X$ is a topological space and $S$ is a subset of $X$ then the subspace topology is defined by taking

  • the open sets of $S$ to be every set of the form $U \cap S$ where $U$ is an open set in $X$

  • the closed sets of $S$ to be every set of the form $F \cap S$ where $F$ is a closed set in $X$

For the rational numbers, closed intervals defined by irrational endpoints are the same as open intervals: if $\alpha < \beta$ are irrational numbers then

$$ [\alpha, \beta] \cap \mathbb Q = (\alpha, \beta) \cap \mathbb Q.$$

This is because if $\alpha \le x \le \beta$ and $x$ is rational then $x \ne \alpha$ and $x \ne \beta$ (since $\alpha, \beta$ are irrational and $x$ is rational) so $\alpha < x < \beta$.

This means that the set $ [\alpha, \beta] \cap \mathbb Q = (\alpha, \beta) \cap \mathbb Q $ is both open and closed in the subspace topology of $\mathbb Q$. That is, it is both an open set intersected with $\mathbb Q$ and a closed set intersected with $\mathbb Q$.

For $S = [0,1] \cap \mathbb Q$ we have

$$\left[0, \frac1\pi - \frac1n \right] \cap S = \left[0, \frac1\pi - \frac1n \right) \cap S = \left(-1, \frac1\pi - \frac1n \right) \cap S. $$

Thus the set $\left[0, \frac1\pi - \frac1n \right] \cap S$ is both open and closed in the subspace topology.

Trevor Gunn
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  • Sir first of all thanks for answering. Please would you like to provide some more "basic" examples and please also discuss about compact sets in subspace topology. – Akash Patalwanshi Feb 27 '18 at 04:34
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    @AkashPatalwanshi Well, the examples don't get too much simpler than intervals in $\mathbb R$ intersecting a subset of $\mathbb R$. Perhaps you should spend some time reviewing the basic concepts of topology: open sets, closed sets, metric spaces. – Trevor Gunn Feb 27 '18 at 04:48
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    @AkashPatalwanshi - instead of demanding that people just give you more examples, you will find it much more productive to be explicit about what you don't understand from Trevor Gunn's answer. Nobody is going to be interested in playing "guess what I want" with you. Trevor has given you the definitions you asked for (except for compactness in a subspace, but that is just the normal definition of compactness, using the subspace topology) and explained why your example looks confusing, but is still correct. What more are you after? – Paul Sinclair Feb 27 '18 at 05:32
  • Thank you so much to all of you. Can anyone explain me this, "The subspace topology of the natural numbers as a subspace of $\mathbb{R}$ is the discrete topology" – Akash Patalwanshi Feb 27 '18 at 12:57
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To answer your question about compact sets in subspace topology, here's an interesting result. If $K \subset Y \subset X$, $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$. (Theorem 2.33, Rudin's Principles of Mathematical Analysis). The proof given there applies to a general topological space, not just metric spaces. So it makes sense to say that a topological space is/is not compact in its own right, irrespective of its embedding space.

Anu
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