Probability of $B_t < 0$ if $B$ is Brownian Motion Let's consider Brownian motion $B_t$. We assume that $B_0 = 0$.
I am to show that 
$$P(\inf{ \{t>0: B_t <0 \}} = 0) = 1.$$
It seems pretty obvious. I don't know how to start the proof however. I would appreciate any hints or tips.
 A: Here are two methods that don't use any particularly advanced facts about B.M. (which you prefer will depend on what things you know about B.M.)

(Via the reflection principle)
Let $M(t) = \sup_{0 \leq s \leq t} B(t)$. The Reflection principle states that $$\mathbb{P}(M(t) \geq a) = 2 \mathbb{P}(B(t) > a) = 2 - 2\Phi(\frac{a}{\sqrt{t}})$$
where $\Phi$ is the cdf of a standard Gaussian distribution. Since $-B(t)$ is also a Brownian motion, it is enough for us to compute $\mathbb{P}(\sup_{t > 0} B(t) > 0)$. But then we have for any $T$,
$$\mathbb{P}(\sup_{t > 0} B(t) > 0) \geq \mathbb{P}(M(T) > 0) = \lim_{n \to \infty} \mathbb{P}(M(T) \geq \frac1n) = \lim_{n \to \infty} 2 - 2 \Phi(\frac1{n \sqrt{T}}) = 1$$

(Via Blumenthal's $0$-$1$ law)
Let $\mathcal{F}_t$ be the filtration generated by your Brownian motion. Blumenthal's $0$-$1$ law tells us that if $A \in \mathcal{F}_{0+} = \cap_{s > 0} \mathcal{F}_s$ then $\mathbb{P}(A) \in \{0,1\}$. 
Let $A = \cap_{n \geq 1} \{\inf_{0 \leq s \leq n^{-1}} B_s < 0\}$ so that $A \in \mathcal{F}_{0+}$. Note that $$\mathbb{P}(A) = \lim_{n \to \infty} \mathbb{P}(\inf_{0 \leq s \leq n^{-1}} B_s < 0) \geq \lim \inf \mathbb{P}(B_{n^{-1}} < 0) \geq \frac{1}{2}$$
So by Blumenthal's $0$-$1$ law, $\mathbb{P}(A) = 1$. This is again a stronger result than desired.
A: This may be criticized as an overkill but a one line answer to your question comes from the Law of Iterated Logarithm for BM: $\lim \inf_{ t\to 0} \frac {B_t} {\sqrt {2t\log\, \log\, (\frac 1 t)}} =-1$ a.s.. This implies that every interval $(0,\delta)$ contains points $t$ with $B_t <0$. 
