Here is Prob. 11, Sec. 24, in the book Topology by James R. Munkres, 2nd edition:

If $A$ is a connected subspace of (a topological space) $X$, does it follows that $\mathrm{Int} A$ and $\mathrm{Bd} A$ are connected? Does the converse hold? Justify your answers.

I know that the closed interval $[0, 1]$, for example, is a connected subspace of $\mathbb{R}$, and so is its interior, which is the open interval $(0, 1)$, but not its boundary $\{ 0, 1 \}$.

Similarly, the set $$ A \colon= \left\{ \ x \times y \in \mathbb{R} \times \mathbb{R} \ \colon \ x^2 + (y - 1)^2 \leq 1 \ \right\} \cup \left\{ \ x \times y \in \mathbb{R} \times \mathbb{R} \ \colon \ x^2 + (y + 1)^2 \leq 1 \ \right\}, $$ which geometrically speaking is the union of the interiors and the boundaries of two circles of unit radius centered at the points $(0, 1)$ and $(0, -1)$ in the plane, is a connected subspace of $\mathbb{R}^2$, and so is its boundary; however its interior, which is the set $$ \left\{ \ x \times y \in \mathbb{R} \times \mathbb{R} \ \colon \ x^2 + (y - 1)^2 < 1 \ \right\} \cup \left\{ \ x \times y \in \mathbb{R} \times \mathbb{R} \ \colon \ x^2 + (y + 1)^2 < 1 \ \right\}, $$ and is thus the union of two disjoint open sets in $\mathbb{R}^2$, is not connected.

Finally, the set $A \colon= \mathbb{Q} \cup (-\infty, 0)$, for example, is a disconnected subspace of $\mathbb{R}$: the sets $A \cap (-\infty, \sqrt{2} )$ and $A \cap ( \sqrt{2}, +\infty)$ constitute a separation of $A$.

However, the boundary of $A$ equals $[0, +\infty)$, and the interior of $A$ equals $(-\infty, 0)$, both of which are connected.

Please refer to this page.

Now my question is as follows:

Can we give any example of a subspace $A$ of a topological space $X$ such that $A$ is connected but neither $\mathrm{Int} A$ nor $\mathrm{Bd} A$ is connected?

  • $\begingroup$ the boundary of $\mathbb{Q} \cap (-\infty,0)$ equals its closure, thus $(-\infty, 0]$. $\endgroup$ – Henno Brandsma May 20 '18 at 11:32

With $X=\Bbb R^2$, try the (even star-shaped) subspace$$A=\{\,(x,y)\in\Bbb R^2\mid x^2+y^2\le 1\lor 1\le x\le 2\,\}.$$

The interior of $A$ is $$\operatorname{Int}A=\{\,(x,y)\in\Bbb R^2\mid x^2+y^2< 1\,\}\cup \{\,(x,y)\in\Bbb R^2\mid 1< x< 2\,\}$$ and these two parts are separated by the line $x=1$. The boundary is $$\operatorname{Bd}A=\{\,(x,y)\in\Bbb R^2\mid x^2+y^2= 1\lor x=1\,\}\cup \{\,(x,y)\in\Bbb R^2\mid x=2\,\}$$ and these two parts are separated by the line $x=\frac32$.

  • $\begingroup$ I'm forgetting math at a tremendous pace, but I discarded this idea because I thought $\{(x,y)\in \mathbb{R}^2 | 1\lt x\lt 2 \}$ was not part of the interior. If you compute the interior as the complement of the closure of the complement, wouldn't those points be in the closure of the complement? I was trying to come up with some intermediate space $H$ between $X$ and $A$ such that the interior of $A$ as a subspace of $H$ had this property, like, maybe take some thickening of $A$ away from the line segment. $\endgroup$ – Callus May 20 '18 at 12:44

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