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When we study Hausdorff space, we can have the following two results:

Cor. 173

If $f,g: X \to Y$ are continuous and $Y$ is $T_2$, then $\{x \in X\mid f(x) = g(x)\}$ is closed in $X$.

Cor. 174

If $f,g:X \to Y$ are continuous and $Y$ is $T_2$ and $\{x \in X\mid f(x)=g(x)\}$ is a dense subset of $X$, then $f=g$.

In fact, these two statements can be used to characterize Hausdorff.

For 173, we have characterization 1:

if $Y$ is not Hausdorff, there always exists a topological space $X$ and two continuous functions $f$ and $g$, such that the set $\{x | f(x) = g(x)\}$ is not closed.

Thus, “a space Y is Hausdorff if and only if for every topological space X and for any continuous maps from X to Y, the set $\{x | f(x) = g(x)\}$ is closed.” is a characterization of Hausdorff Space.

174 is similar, we have characterization 2:

$Y$ is Hausdorff, if and only if whenever we have $f$ and $g$ continuous from any space $X$, and $\{x | f(x) = g(x)\}$ is a dense subset of $X$, then $f=g$”

My question is: How to prove these, i.e., How to find examples to prove the other direction of characterization 1,2?

Characterization 2 already has an answer here: Does this property characterize a space as Hausdorff?

Henno Brandsma
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Hamilton
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    Please write all mathematical results in the text itself instead of posting a picture. This makes it more searchable and is easier to read. (With more than three years on this site, you are not a "beginner" anymore :-) – PatrickR Oct 21 '21 at 07:16
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    @PatrickR I followed up on your request. Sorry to encourage laziness this way but it benefits the site.. – Henno Brandsma Oct 21 '21 at 15:01
  • @HennoBrandsma You are too nice! I was expecting the asker would do it. – PatrickR Oct 21 '21 at 20:47
  • @Beginner We have been talking about it on Meta: https://math.meta.stackexchange.com/questions/34216 Hope you can do that in the future, especially since the mathematical contents of your questions is good. That will benefit everyone. Thanks in advance! – PatrickR Oct 21 '21 at 20:49
  • https://math.meta.stackexchange.com/questions/34121 – PatrickR Oct 21 '21 at 20:59
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    @PatrickR, Thank you for your patience, I would not put the central part of the problem just in a picture in the future, and I understand that I can only use images to demonstrate some supportive/auxiliary content, in order to make my question searchable, accessible... Thank you for all of your effort to make this community better. – Hamilton Oct 21 '21 at 22:02

1 Answers1

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For any directed set $I$ (such as we use in nets) we define $X(I):=I \cup \{\infty\}$ where $I$ consists of isolated points and a basic neighbourhood of $\infty$ is of the form $\{\infty\} \cup \{i \in I: \mid i \ge i_0\}$ where $i_0 \in I$ is arbitrary. It's easily shown that this defines a topological space. In it$ I$ is dense and non-closed.

Then if $Y$ is not Hausdorff we have a net $(x_i)_{i \in I}$ in $Y$ that converges to two distinct points $p\neq q \in Y$.

Then defining $f,g: X(I) \to Y$ by $f(i)=x_i = g(i)$ and $f(\infty)=p, g(\infty)=q$ we easily check that both are continuous maps (due to the net converging to both points), and $I=\{x\mid f(x)=g(x)\}$ is both not a closed set and also dense despite $f \neq g$. So both claims are shown at the same time.

Henno Brandsma
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