In my general topology textbook they made the following question:
Is it true that every subspace of a connected space is connected?
I answered affirmative and I came up with the following proof:
My proof
Let $(X,\tau)$ be a topological space and let $(Y,\tau_y)$ be a subspace of $(X,\tau)$.
Let's assume that $(X,\tau)$ is connected: this means that the only clopen subsets of $X$ are $X$ and $\emptyset$.
Let $A \neq X$ and $A \neq \emptyset$. Let $A \in \tau$, then $X\setminus A \notin \tau$.
Because $A \in \tau$ and $X\setminus A \notin \tau$, then we know that $A \cap Y \in \tau_y$ and $(X\setminus A)\cap Y \notin \tau_y$. We know that $(X\setminus A)\cap Y = Y\setminus A$.
We also know that $Y\setminus (A \cap Y) = Y\setminus A = (X\setminus A)\cap Y \notin \tau_y$.
So if $A \cap Y \in \tau_y$ then $Y\setminus (A \cap Y) \notin \tau_y$, which means that the set cannot be clopen, thus proving that the only clopen sets are $Y$ and $\emptyset$, and that $(Y,\tau_y)$ is also connected?
Is this proof correct? Because I searched online and some people claimed that the answer to the question is negative, yet I "proved" affirmative. What mistake did I do?
Best Answer
$A\cap Y$ can be in $\tau_y$ and $(X-A)\cap Y$ in $\tau_y$ also, consider $\{0,1\}\subset \mathbb{R}$ and $A=(-1/2,1/2)$
$A\cap \{0,1\}=\{0\}$ and $(X-A)\cap\{0,1\}=\{1\}=(1/2,3/2)\cap\{0,1\}$.
There can exists $A'\in \tau$ with $A'\cap Y=A\cap Y$.