Is regularity preserved in subsets of regular spaces?

[ice109]

i've texed up three proofs in from elementary topology. can someone please check them?

actually i'll just retype them here for convenience

8.2.5

Let $f: X_{\tau} \rightarrow Y_{\nu}$ be continuous and injective. Also let $Y_{\nu}$ be Hausdorff.

Prove : $X_{\tau}$ is Hausdorff.

Proof : Pick $y_1$ and $y_2$ in $f(X)$. By injectivity of f there exist $x_1 =$
$f^{-1}(y_1)$ and $x_2 = f^{-1}(y_2)$ such that they are both unique. $f(X)$ is Hausdorff so by Theorem 1 there exist disjoint open neighborhoods $U$ and $V$ of $y_1$ and $y_2$ respectively. Then $f^{-1}(U \bigcap V) = f^{-1}(U) \bigcap f^{-1}(V) = \emptyset$ and by continuity $f^{-1}(U)$ and $f^{-1}(V)$ are open. Finally by definition of $f^{-1} : x_1 \in f^{-1}(U)$ and $x_2 \in f^{-1}(V)$ which, as stated previously, are two open disjoint sets in $X_{\tau}$. Hence $X_{\tau}$ is Hausdorff.\\



Comments:

Injectivity is necessary. Take for example $X = \{a,b\}$ and $Y = \{a\}$ and to be $f: X_{\mathcal{I}} \rightarrow Y_{\mathcal{I}}$. Explicitly $f(\{a,b\}) =\{a}\}$. f is continuous, $Y_{\mathcal{I}}$ is obviously Hausdorff and $X_{\mathcal{I}}$ is obviously not.\\

This does not prove that Hausdorff is a strong topological property because we have proven a stronger converse. To prove that Hausdorff is a strong topological property we would have to have proven that $f: X_{\tau} \rightarrow Y_{\nu}$ continuous, not necessarily injective, and $X_{\tau}$ Hausdorff implies $f(X)$ Hausdorff.

8.2.7

Let $X_{\tau}$ be $T_1$ and $A \subseteq T$ and $x \in A'$.\\

Prove : Any neighborhood of x intersects \textit{A} in infinitely many points.

Proof : Assume that there exists a neighborhood of x, in $X_{\tau}$ that intersects \textit{A} in only finitely many points to derive a contradiction. Let $N_x$ be such a neighborhood. $N_x$ is $T_1$ by Theorem 1. Hence we can separate x from all points in $N_x$ by other neighborhoods. Since there are finitely many points in $N_x$ there are finitely many such neighborhoods. Let \textit{N} be the intersection of those neighborhoods. $N -\{x\}$ therefore is itself a non-trivial open neighborhood of \textit{x} which does not intersect \textit{A}. This contradicts that x is a limit point of \textit{A} and therefore any neighborhood of x intersects \textit{A} in infinitely many points.



\textbf{8.3.4}\\

Let $A \subseteq X_{\tau}$ and $X_{\tau}$ be regular.\\

Prove : A is regular.

Proof : Pick $A_1 \subseteq A$, $A_1$ closed in the subspace topology, and $x \in A - A_1$. Then $A_1 = B$ for some $B \in \tau$ and we can find two open sets $N_B$ and $N_x$ by the the regularity of $X_{\tau}$ which are disjoint. $N_B \bigcap A$ and $N_x \bigcap A$ are two disjoint sets in A which contain $A_1$ and x respectively. Therefore A is regular.


Theorem 1 : A subspace of a $T_i$ space for $i \leq 2$ is $T_i$.

 

[futurebird]

I'm not an expert at all... but in 8.2.5 shouldn't you say "pick $$y_{1} \neq y_{2}$$ in $$f(X)$$?

 

[ice109]

I'm not an expert at all... but in 8.2.5 shouldn't you say "pick $$y_{1} \neq y_{2}$$ in $$f(X)$$?

yea you're right

 

[mrandersdk]

yea you're right

When you want to prove that X is hausdorff I think it is nicer to say: Pick $x_1 \neq x_2$ in X, and then do as you do. It is because when it is X you want to prove is hausdorf I think you should start there, because you have to prove it for any two x'es in X.