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EDIT. As requested, I am extending the answer, including the relevant definitions. To do so, I have also re-arranged the answer somewhat. I am also now including the notions of surjective span and sem …

As pointed out by Jeff, the notion you define may not really be what you are after, since indecomposable continua are not 'indecomposable' in your sense. However, we can ask: Is there a nontrivial co …

I realise this is quite an old question, but here is an alternative to the already existing elegant answers, which shows that Hausdorff can be replaced by $T_1$: Proposition. No countable $T_1$ sp …

I believe the question whether $X$ can be chosen to be Hausdorff was left open by both existing answers. The solution is provided by the H-closed spaces of Henno Brandsma's comment. I shall answer the …

Any indecomposable continuum has the property you desire. (A continuum is indecomposable if it cannot be written as a union of two proper subcontinua.) One way to see this is that any indecomposabl …

I think your assumption should include that the set contains at least two points, otherwise there is a trivial example ... Eric already mentioned the pseudo-arc, which is of course a perfect answer. …

The question has been answered in the comments, but just for the record: The homeomorphic image of a disc (even embedded in $\mathbb{R}^n$ for $n>2$, or indeed in any metric space) cannot have Hausdor …

Let $K$ be the boundary of your circle domain $\Omega$. Let us suppose that every point of $K$ is accumulated on by a sequence of (pairwise different) circle components. Such an example is easy to c …

In the broad sense in which you state your question, and if you are looking for rigorous results, then the answer is surely no. Keep in mind that the existence of the Lorenz attractor was only prove …

There are indeed quite a number of definitions of end-points, or of terminal points (the terms are sometimes used interchangeably, sometimes not) of a continuum $X$, as I discovered recently when I wa …

I just stumbled upon this old question, and thought I would add a simple and natural example, which is $G_{\delta \sigma}$ but neither $G_{\delta}$ nor $F_{\sigma}$. Consider $f\colon \mathbb{C}\to\ma …

I know this is an old question that already has an excellent answer (although it may apply only to compact sets). However, let me respond to the original question, concerning the following results: …

I am interested in a reference for the following fact (or a similar result). PROPOSITION. Let $X$ denote the set of real null sequences; i.e., the set of $(a_n)_{n=0}^{\infty}$ with $a_n\to 0$, with …

Let $f$ be the complex polynomial $f(z) = z^2 - 1$. Its Julia set $J(f)$ is the set of non-equicontinuity of the iterates; i.e., the set of points whose orbit under $f$ is not stable under perturbatio …

Here is one possible elementary argument (somewhat inspired by my paper with Shen in the Monthly, "The exponential map is chaotic"), which avoids any mention of the classification of Fatou components, …