# General topology – A suslinian and non-suslinian continuum?

There is an apparent contradiction in the literature that I would like to address …

continuum means compact metizable connected with more than one point.

A continuum is Suslinean if each collection of discontinuous sub-discontinues in pairs is countable.

In the following example, Author A build a continuum $$Y: = X / sim$$ which is the closing of a ray (a one-to-one image of $$[01)[01)[01)[01)$$) so that the radius is in first category $$Y$$. In other words, the radius and its complement are both dense in $$Y$$.

It's clear that $$X$$ is Suslinean, so $$Y = X / sim$$ is also Suslinean.

On the other hand, since $$Y$$ contains a dense ray of first category, there is a sequence of disjoint arcs in pairs in $$Y$$ which converges towards $$Y$$ in the distance Hausdorff. By a theorem of Author B, $$Y$$ is non-Suslinean.

Question 1: Is there really a contradiction? I must mention that both articles have appeared in good journals and that both authors are well respected in this field.

Question 2: Is the example correct? I found some typos, for example $$A_n$$ should be $$C_n cup bigcup …$$ and $$overline {z_1 z_1}$$ should be $$overline {z_1 z_2}$$but otherwise it looks good.

If the answer to these questions is YES, then I will have to look for a flaw in the proof of the author B.