Consider $\mathbb{R}$ with its standard topology and let $A=(0,1)\cup\{2\}$ subset of $\mathbb{R}$. Determine all open sets in the subspace $A$ of $\mathbb{R}$. This problem is in Topology by Djugundji but I cannot check my answer if it right since it does not have solution. Any help would be appreciated..


Any $X \subseteq A \cap (0,1)$ that is open in $\mathbb R$ is open in $A$ and for each of these $X \cup \{2\}$ is also open in $A$. Every open subset of $A$ has one of these two forms. I leave the easy proof of this to you (you just have to unwind the definition of the subspace topology).

  • $\begingroup$ The question is to determine so I have to enumerate all open sets? Can you give me a figure how many are open sets? Thanks a lot. $\endgroup$ – Bayoy Mar 11 '16 at 12:07
  • $\begingroup$ Why is $X \cup [2]$ open ? (I assume that {2} is the closed set $[2]$ - so why is the union of an open and closed set open ? $\endgroup$ – Tom Collinge Mar 11 '16 at 12:11
  • $\begingroup$ @TomCollinge $\{2\}$ is open in the subspace topology, because $\{2\} = (1.5,2.5) \cap A$. $\endgroup$ – Stefan Mesken Mar 11 '16 at 12:13
  • $\begingroup$ @Bayoy You can't enumerate all open sets (there are precisely $2^{\aleph_0}$ many of them). I'm pretty sure 'determine' means 'describe' in this context. $\endgroup$ – Stefan Mesken Mar 11 '16 at 12:14
  • $\begingroup$ OK, Thanks. I missed the subtlety of the subspace topology. $\endgroup$ – Tom Collinge Mar 11 '16 at 12:14

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