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Given the set $ X=\Bbb N \cup \{-1/n|n\in \Bbb N\} $ whose topology $\tau$ is generated by the basis elements $(-1,b)\cup (c,d) \cup (a,\infty)$, I want to decide if $(X,\tau )$ is homeomorphic to the topological space $(X,T_x)$ where $T_x$ is the subspace topology induced by the standard topology on $\Bbb R$.

Here is my attempt:

First of all the subspace topology has as its open sets unions of 1 point sets , so we need to decide if there exists a continuous bijective function whose inverse is also continuous which maps one point sets $\{x\}$ where $x\in \Bbb N \cup \{-1/n|n\in \Bbb N \}$ onto either $(-1,b),(a,\infty)$or (c,d) and whose preimage will be open in X.

Well the subspace topology is the same as the discrete topology in this case and so no matter what the function is it will always have as its preimage an open set as all sets are open in this case.

Does that seem right ? I'm not really sure if I'm expressing it lucidly enough if it is correct so any advice, perhaps, to tidy my argument would be welcome also .

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  • $\begingroup$ You should also consider $f^{-1}$, whether it is continuous. Here is a different hint. If $E=\{2n|n\in \Bbb N\} \cup \{-1/n|n\in \Bbb N\}$ then $E$ is open in the discrete topology, but not in the other topology that you consider. Since every subset is open in the discrete topology, the two spaces are not homeomorphic. (Any bijection you take will match $E$ with a set that is necessarily open in the discrete topology.) Note also that the topology with that basis is $T_1$ but not $T_2$, i.e. not Hausdorff. $\endgroup$ – Mirko Jul 26 '19 at 12:54
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    $\begingroup$ No, those weird base elements make that space compact. $\endgroup$ – William Elliot Jul 26 '19 at 12:57
  • $\begingroup$ @WilliamElliot Your comment made me realize the basic elements should likely be $[-1,b)\cup (c,d) \cup (a,\infty)$ rather than $(-1,b)\cup (c,d) \cup (a,\infty)$ since $-1$ is not a member of any $(-1,b)\cup (c,d) \cup (a,\infty)$, even if $-1\in X$. $\endgroup$ – Mirko Jul 26 '19 at 13:02
  • $\begingroup$ @Mirko ah okay, I was only considering it in one direction ( for f) and not the other (for f inverse) . Also I think I was focusing too much on using the definition (i.e. with contiuous functions preimages etc) instead of focusing on the main point of homeomorphisms which is that they have direct correspondence between open sets . Thanks for your wisdom :) $\endgroup$ – excalibirr Jul 26 '19 at 13:02
  • $\begingroup$ @Mirko that was a typo , it was supposed to be $[-1,b)$ $\endgroup$ – excalibirr Jul 26 '19 at 13:03
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Edit: I apologize, I have slightly misread the definition of $X$. I have corrected the proof accordingly.

$X$ is not compact in the subspace topology: consider the open cover

$$\mathscr{A}=\left\{\left(\frac{-1}{n},\frac{-1}{n+2}\right)\cap X:n\in \mathbb{N}\right\}$$ together with the open sets $\{-1\}$ and $\{n\}$ for each $n\in \mathbb{N}$.

Each element $A\in \mathscr{A}$ contains only one element of $X$, so no finite subcollection of $\mathscr{A}$ covers $X$.

But $X$ is compact in your topology. Any open set $A$ containing $1$ contains infinitely many points of $X$ - for some $N$, for all $n\geq N$, $\frac{-1}{n}$ is contained in $A$. Additionally, $A$ also contains infinitely many points of $X$ which are positive integers. Then there are only finitely many points in $X$ left: finitely many negative fractions, and finitely many positive integers. In any open cover $\mathscr{A}$, pick an element $A\in \mathscr{A}$ which contains $1$; $A$ covers all but finitely many elements of $X$. So simply pick one element of $\mathscr{A}$ to cover each one.

So the two spaces are not homeomorphic.

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  • $\begingroup$ Why is {1} open? Your cover does not include points like 2 and 3. $\endgroup$ – William Elliot Jul 27 '19 at 0:47
  • $\begingroup$ @William Elliot; I misread the definition of X; I have fixed it above. $\endgroup$ – Laurel Turner Jul 27 '19 at 16:09

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