# Path connected

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• Oct 17th 2011, 04:03 PM
dwsmith
Path connected
If X and Y are path connected, show that XxY is also path connected.

For a fixed $x_0\in X$ and $y_0\in Y$
Let A be the set of all the points that join $x_0\times y_0$ by a path in $U\times V$.

Since X and Y are both path connected, there exist $\epsilon >0$ such that $B(x,\epsilon) \ \text{and} \ B(y,\epsilon)$, where x is in $\pi_1(A)$ and y is in $\pi_2(A)$, contained in open sets $U\subset X$ and $V\subset Y$, respectively.

Let $x_1\in B(x,\epsilon)$ then a path can be connected from $x_1$ to x to $x_0$.

Similarly for y

How can now show XxY is path connected?

How can I translate this to the product topology?

Can I say:

$x_1\times y_1\in B(x\times y,\epsilon)\subset A\subset U\times V$. Therefore the exist a path from $x_1\times y_1$ to $x\times y$ to $x_0\times y_0$????
• Oct 19th 2011, 08:51 PM
MonroeYoder
Re: Path connected
Suppose X and Y are path connected. By definition, for any $x_1,x_2\in X$ there is a continuous function $f:[0,1]\rightarrow X$ with $f(0)=x_1$ and $f(1)=x_2$

Now let $(x_0,y_0)$ and $(x_1,y_1)$ be two points in $X\times Y$and let $f$ be the path in $X$ from $x_0$ to $x_1$ and $g$ be the path in $Y$ from $y_0$ to $y_1$. Define $h:[0,1]\rightarrow X\times Y$ by $h(s) = (f(s),g(s))$. Then $h(0)=(x_0,y_0)$ and $h(1)=(x_1,y_1)$.

You should know a theorem that tells you $h$ is continuous since it maps into a product space and each coordinate is given by a continuous function(Thm 19.6 in Munkres)