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Question: Let X be a Hausdorff space and Y be a subset of X. Then, Y with the subspace topology is a Hausdorff space.

This is what I did, can someone verify this and let me know if I am correct or wrong? Also, kindly let me know if my proof need some changes or modifications due to bad notations.

Any help will be greatly appreciated.

Math_Is_Fun
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    Looks correct. The one pedagogical point I would make is to note explicitly that $U$ and $V$ are open in $X$. That's clear, but perhaps not immediately obvious since you're working in two different topologies. – Robert Shore Mar 06 '20 at 00:54
  • It's all correct. – Berci Mar 06 '20 at 00:55

1 Answers1

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Everything that needs to be there is there, so it's a valid proof. My only comments are about the style.

  • The line where you recall the definition of the subspace topology on $Y$ is out of place. You've already used this definition once; either you should state it at the top before you use it the first time, or leave it out altogether.

  • “Hence a set containing $x$ in $Y$ is $U' = U \cap Y$, which is open in $Y$” might read better if it were written as “Hence $U' = U \cap Y$ is open in $Y$ and contains $x$.” You're defining $U'$ in this sentence, so I feel like that definition should come first, and its properties later.

Otherwise good job!

Matthew Leingang
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  • Thanks a lot, will take a note of your suggestions and use it to make even better proof. Appreciate your help!! – Math_Is_Fun Mar 06 '20 at 04:03