I'd like a check about this exercise.

Let $\tau$ be the topology so defined $\tau=\bigl\{A \subset \mathbb{R}\,|\, (\forall x \in A \cap \mathbb{Z})( \exists \epsilon >0): (x-\epsilon,x+\epsilon) \subset A\bigr\}$

(i) Is $\tau$ finer or coarser than the euclidian topology?

(ii) Find $\operatorname{Int}[0,+\infty), \operatorname{Int}[0,1], \operatorname{Int}\mathbb{Z}$

(iii) Find $\overline{[0,1]}$

(iv) Is $\mathbb{R},\tau$ an Hausdorff space?

(v) Is $\mathbb{R}$ compact

(vi) Let $f\colon(\mathbb{R},\tau_e) \rightarrow (\mathbb{R},\tau_e)$ be a continuous function such that $f(\mathbb{Z}) \subset \mathbb{Z}$.

Show that $f\colon(\mathbb{R},\tau) \rightarrow (\mathbb{R},\tau)$ is continuous.

(i) $\tau_e$ is finer than $\tau$ because $\tau \subset \tau_e$. In fact, every open set in $\tau$ is open in the standard euclidian topology. I also have that $\tau_e \subset \tau$, so the two topologies are equivalent.

(ii)$\operatorname{Int}[0,1]=\emptyset$, since the only integers in this set are {$0,1$}, but if $0$ is included, then $(0-\epsilon,0+\epsilon) \notin [0,1]$

With the same argument, $\operatorname{Int}[0,+\infty)=(0,+\infty)$

$\operatorname{Int}\mathbb{Z}=\emptyset$ because every open set which contains integer points, automatically contains also real points.

(iii) $[0,1]$ is closed since $(\infty,0) \cup (1,+\infty)$ is an open set. So $\overline{[0,1]}=[0,1]$.

(iv)Yes, it's an Hausdorff space since for every different points $x,y \in \mathbb{R}$ I can find two open neighborhood $U,V$ such that $U \cap V = \emptyset$.

(v)It's not compact: if I take the open cover $\mathcal{R}= \cup_{i \in \mathbb{N}}A_i$, $A_i=(i,i+2)$, I can't take a finite subcover.

(vi) Since $f$ is continuous (with the euclidean topology), then for every open set $U \in \tau_e$: $f^{-1}(U) \in \tau_e \subset \tau$. So $f:(\mathbb{R},\tau) \rightarrow (\mathbb{R},\tau)$ is continuous.

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    $\begingroup$ I don't understand the definition of $\tau$; what is $\epsilon$? $\endgroup$ – José Carlos Santos Aug 30 '17 at 12:26
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    $\begingroup$ If there is supposed to be an $\exists \epsilon$ right before the interval in the definition of $\tau$, then your answers are mostly wrong. Note in particular that $(1,2)$ is open in $\tau$. $\endgroup$ – Mees de Vries Aug 30 '17 at 12:29
  • $\begingroup$ Sorry, I edited my answer $\endgroup$ – VoB Aug 30 '17 at 12:31
  • $\begingroup$ One way to see this topology is as the smallest topology such that $f:(\mathbb R,\tau)\to X$ is continuous for a metric space $X$ if and only if $f:\mathbb R\to X$ is continuous at every point of $\mathbb Z$ (using the $\epsilon-\delta$ definition of 'continuous at a point.' $\endgroup$ – Thomas Andrews Aug 30 '17 at 12:47
  • $\begingroup$ Too many questions. $\endgroup$ – Jack Aug 30 '17 at 13:07

First of all, you're misunderstanding $\tau.$ Putting the set-builder notation into words, the elements of $\tau$ are the subsets $A$ of $\Bbb R$ which contain euclidean neighborhoods of each of their integer elements. However, this does not mean that they contain euclidean neighborhoods of all of their elements. For example, taking any $x\in\Bbb R\setminus\Bbb Z,$ we have that $\{x\}\in\tau,$ but $\{x\}\notin\tau_e.$

Second of all, you've made correct arguments for determining $\mathrm{Int}[0,1],$ but you've drawn an incorrect conclusion. You're right that $0,1\notin\mathrm{Int}[0,1],$ but those are the only elements of $[0,1]$ you've proved (and indeed, the only ones that you can prove) not to be in $\mathrm{Int}[0,1]$.

Your argument for (iv) needs to be fixed, as it currently says: "Yes, it's a Hausdorff space, because it's a Hausdorff space."

Your argument for (v) needs to be fixed, because $\{A_i:i\in\Bbb N\}$ does not cover $\Bbb R.$ That's an easy fix, though.

Your approach for (vi) needs to be fixed. You must show that for every $U\in\tau,$ we have $f^{-1}[U]\in\tau.$

Let me know if you have trouble with any of these, and I'll see what I can do to get you "unstuck."


As pointed out in the comments, some are wrong, but more importantly, you seem to have the wrong intuition about $\tau$. The first answer that is wrong is (i).

From the definition of $\tau$ it follows that $(1,2)\in \tau$ (Because $(1,2)\cap \mathbb Z=\emptyset$ and you take $x$ from an empty set in the definition).

In fact, it can be shown that every subset of $\mathbb R\setminus \mathbb Z$ is in $\tau$.

  • $\begingroup$ thanks for the answer ! I just edited ;) $\endgroup$ – VoB Aug 30 '17 at 12:49

Here's my solution of (vi): Let $f\colon\mathbb{R}\longrightarrow\mathbb R$ be a function which is continuous with respect to the usual topology. You want to prove that it is continuous with respect to $\tau$. Let $x\in\mathbb R$ and let $U$ be a neighborhood of $f(x)$; you want to prove that $f^{-1}(U)$ is a neighborhood of $x$. There are two possibilites:

  1. $x\in\mathbb Z$: then $f(x)\in\mathbb Z$ and therefore $U\supset\bigl(f(x)-\varepsilon,f(x)+\varepsilon\bigr)$ for some $\varepsilon>0$. Since $\bigl(f(x)-\varepsilon,f(x)+\varepsilon\bigr)\in\tau_e$, $f^{-1}\bigl(f(x)-\varepsilon,f(x)+\varepsilon\bigr)\supset(x-\delta,x+\delta)$, for some $\delta>0$. Therefore, $f^{-1}(U)\supset(x-\delta,x+\delta)$ and so $f^{-1}(U)$ is a neighborhood of $x$.
  2. $x\notin\mathbb Z$: then $\{x\}$ is a neighborhood of $x$ and therefore $f^{-1}(U)(\supset\{x\})$ is also a neighborhood of $x$.

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