00:01
Hi, in this question, to show that x is a host of space, we need to demonstrate that for any two distinct points x and z and x, there exist neighborhoods of x and z that are disjoint from each other.
00:13
So, they give one that for each x not equal to z, then x, there is a continuous function, f is to x to or comma mu such that f of x is equal to 1 and f of z is equal to 0.
00:42
We can use this information to construct the neighborhoods.
00:46
Let us take u is equal to f inverse of minus 1 by 2 comma 3 by 2 and v is equal to f inverse of minus 1 by 2 comma 1 by 2.
01:02
So, here f inverse denotes the preimage of a set under the function f.
01:12
Since f is continuous, u and v are open sets in x.
01:28
Additionally, u contains x and v contains z.
01:37
Now, let us show that u and v are distinct...