Is $A=(0,1]$ a closed or open set?

I think it's not an open set because it is not a subset of its interior points. Mainly, $1\in A$ but $1\not\in A^\circ$.

If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 \in A^c$ but $0\not\in (A^c)^\circ$

  • $\begingroup$ $(0, 1]$ is a semi-open or semi-closed set. $\endgroup$ Mar 23 '19 at 16:18
  • 5
    $\begingroup$ Depends on the topology! $\endgroup$
    – Jakobian
    Mar 23 '19 at 16:21
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    $\begingroup$ Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading. $\endgroup$
    – Simon
    Mar 23 '19 at 16:23
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    $\begingroup$ The answer to your question is no. $\endgroup$ Mar 23 '19 at 16:42
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    $\begingroup$ @Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once. $\endgroup$ Mar 23 '19 at 18:38

Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-\infty,\,0]\cup (1,\,\infty)$, which doesn't contain a neighbourhood of $0$.

  • $\begingroup$ note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set. $\endgroup$ Mar 23 '19 at 18:24
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    $\begingroup$ @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology. $\endgroup$
    – J.G.
    Mar 23 '19 at 18:36

It's important that you specify where you are considering the subset $A$.

If $A \subset X$ with $ X = \mathbb{R}$, J.G. is absolutely right in the usual topology of $\mathbb{R}$.

If $A \subset X$ with $ X = [0,1]$, $A^c = \{0\} $, which is closed in the usual topology, then $A = (0,1]$ is open.

In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.


A set is not a door.

It is not the case that a set is either open or closed. It can also be neither or both.

Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $\mathbb{R}$ with the usual topology. The empty set $\emptyset$ is always both open and closed, no matter what the ambient space is.


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