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The following are simple observations.

Suppose $\mathcal{T}_1,\mathcal{T}_2$ are two topologies on a set $X$ such that $\mathcal{T}_1$ is finer than $\mathcal{T}_2$.

  1. If $( X ,\mathcal{T}_2 )$ is Hausdorff, then so is $( X ,\mathcal{T}_1 )$.
  2. If $( X ,\mathcal{T}_1 )$ is connected, then so is $( X ,\mathcal{T}_2 )$.

The real numbers with the usual topology is both Hausdorff and connected, which leads to the following two questions.

  1. Is there a topology on the real numbers strictly coarser than the usual one, which makes it still Hausdorff?

  2. Is there a topology on the real numbers strictly finer than the usual one, which makes it still a connected space?

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    $\begingroup$ K-topology is finer and connected on $\mathbb R$. en.wikipedia.org/wiki/K-topology $\endgroup$ – user135988 Jun 2 '15 at 22:05
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    $\begingroup$ This previous question gives a necessary condition for a topology to be "minimally Hausdorff" (i.e., no strictly coarser topology is Hausdorff). The reals with the usual topology clearly don't satisfy this condition ($\mathbb{R}$ is homeomorphic to $(0,1)$ which is not closed in $[0,1]$). $\endgroup$ – user642796 Jun 3 '15 at 7:26
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People suggesting that you can't make a coarser, still Hausdorff topology are wrong. Their proofs are invalid, and so is the conclusion. I construct a coarser topology that is still Hausdorff. You can think of this space as adding a point at $+\infty$ and gluing it to 0.

Let $\tau$ be the usual topology of open sets on $\mathbb{R}$. Define $\tau' \subset \tau$ as the collection of sets in $\tau$ that, if they contain 0, contain some interval $(\alpha,+\infty)$. That is, the open sets in our new topology are the old open sets, but, if they contain the point 0, we also require them to be a neighborhood of $+\infty$. By definition $\tau'$ is coarser than $\tau$, and it is immediate that it is closed under finite intersection and arbitrary union. To show it's Hausdorff, say we're trying to separate two points $x$ and $y$. If $x,y \neq 0$, just choose some small intervals around $x$ and $y$, making sure you don't include 0 in either of them. If one of them is zero, just pick intervals as usual, and throw in some $(\alpha,+\infty)$ to the neighborhood of 0, with $\alpha$ large enough not to include the other point.

Someone already provided a working example for a finer topology that's still connected.

Edit: I'll provide some informal rationale about why the compactness of closed intervals is not enough to get what people seem to want, and shouldn't be. Morally, you can think of Hausdorff as meaning "sequences can't converge to more than one point", and a coarser topology as being one in which "more converging happens." The reason that you can't make a coarser Hausdorff topology on a compact space is that there is already lots of converging going on. To get more converging, you would have to make a sequence converge to something new, when it already converges to something else (breaking Hausdorff). The key way we're using compactness here is that we know that "everything that could converge already does"; that's a global property of compactness, and we can't recover it from local compactness, or knowing that our space is a countable union of compact sets, or anything like that. In the construction I gave, we added "more convergence" by saying that sequences that tend to $+\infty$, which ordinarily diverge, should instead be considered converging to 0.

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    $\begingroup$ @Gamamal The open sets are just the usual open sets, with an additional non-vacuous restriction. It therefore has "fewer" open sets, so is coarser. $\endgroup$ – Mike Haskel Jun 2 '15 at 22:33
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    $\begingroup$ @Gamamal No. The underlying set is still $\mathbb{R}$ in the explicit construction. In the "gluing" way of thinking about it, there is not really a new point, since the "new point" at $+\infty$ gets glued to 0. $\endgroup$ – Mike Haskel Jun 2 '15 at 22:34
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    $\begingroup$ @Gamamal : It's like defining a continuous map from $\mathbb R$ to the figure 8 in $\mathbb R^2$ and take the pullback topology. $\endgroup$ – user99914 Jun 2 '15 at 22:36
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    $\begingroup$ It's like equipping $\Bbb R$ with the initial topology with respect to the map from $\Bbb R$ to $X=(-\infty,-1]\cup S^1$ which sends any $x\le0$ to $x-1$, and $x\ge0$ to $e^{i\pi+2\pi i(1-1/(1+x))}$. The initial topology for the map from $\Bbb R$ to $\large\infty$ is even coarser and still Hausdorff. $\endgroup$ – Stefan Hamcke Jun 2 '15 at 22:50
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    $\begingroup$ Stefan might be saying this more rigorously (too lazy to figure it out), but: this solution is like assigning to $\mathbb R$ the topology obtained by identifying it with a subspace of $\mathbb R^2$ given by a circle and a tail. The fact that a 1-1 continuous path from $\mathbb R$ onto such a figure should be pretty intuitive, which is what makes the topology coarser. $\endgroup$ – Dustan Levenstein Jun 2 '15 at 23:01
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For your second question, yes.

Let $\tau$ be the topology on $\mathbb R$ generated by $X\cup \mathbb Q$ or $X$, where $X$ is open in $\mathbb R$ in the usual topology. Then it is connected.

To see this, let $\mathbb R = Y_1 \cup Y_2$ be two disjoint union, where $Y_i \in \tau$. Then either one of them, say $Y_1$, is not open in the usual topology. By definition,

$$Y_1 = X_1 \cap (W_1 \cup \mathbb Q),$$

where $X_1, W_1$ are open (in the usual sense). As this is not open in the usual toplogy, there is $p \in Y_1\cap \mathbb Q$ so that $$(p - \epsilon, p+ \epsilon) \cap Y_1 = (p-\epsilon, p+ \epsilon) \cap \mathbb Q$$

for some $\epsilon >0$. But that is impossible for $Y_2$ to be open in $\tau$.

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